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Example 5 Prove: When $k \geq 3+\sqrt{7}$, we have
$$\sum x \sum \frac{x}{x^{2}+k y z} \geq \frac{9}{1+k} \quad(x>0, y>0, z>0)$$ | Prove that when $k \geq 3+\sqrt{7}$, the equivalent form of equation (3.2.9) is
$$L_{5}(x, y, z)=\left(k^{2}+k\right) f_{0.3}^{(6)}+\left(4 k^{2}-5 k\right) f_{0.4}^{(6)}+k\left(k^{2}-6 k+2\right) f_{1,1}^{(6)}+\left(2 k^{3}-2 k^{2}-3 k+1\right) f_{1.2}^{(6)} \geq 0$$
Since $k \geq 3+\sqrt{7}$, we have
$$\begin{array}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,346 |
Example 6 Let $x>0, y>0, z>0$, prove that
$$3 \leq \sum \frac{y+z-x}{x} \leq \sum\left(\frac{y+z-x}{x}\right)^{2} \leq \sum\left(\frac{y+z-x}{x}\right)^{3} \leq \sum\left(\frac{y+z-x}{x}\right)^{4}$$ | To prove that (3.2.11) is equivalent from left to right, we have:
$$\begin{aligned}
L_{6,1}= & f_{0,2}^{(3)} \geq 0 \\
L_{6,2}= & f_{0,3}^{(6)}+4 f_{0,4}^{(6)}+2 f_{1,1}^{(6)}+f_{1,2}^{(6)} \geq 0 \\
L_{6,3}= & f_{0,4}^{(9)}+4 f_{0,5}^{(9)}+f_{1,2}^{(9)}+2 f_{2,1}^{(9)}+f_{2,2}^{(9)} \geq 0 \\
L_{6,4}= & f_{0,5}^{(12)}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,347 |
Example 7 Let $a, b, c$ be the lengths of the sides of a triangle, and $s, R, r$ be its semi-perimeter, circumradius, and inradius, respectively, then
$$\left(\sum \sqrt{s-a}\right)^{2} \leq\left(4+\frac{r}{R}\right) \frac{\sum b c}{\sum a}$$ | Prove that by letting $\sqrt{s-a}=x, \sqrt{s-b}=y, \sqrt{s-c}=z$, we obtain the equivalent form of (3.2.12):
$$L_{7}(x, y, z)=f_{0,3}^{(8)}-2 f_{0,4}^{(8)}+2 f_{0,5}^{(8)}+2 f_{1,1}^{(8)}-2 f_{1,2}^{(8)}-2 f_{1,3}^{(8)}+4 f_{2,1}^{(8)} \geq 0$$
By combining the terms of the above expression, we get
$$\begin{array}{l}
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,348 |
Example 8 Let $a, b, c$ be the lengths of the three sides of a triangle, $R, r, s$ be the circumradius, inradius, and semiperimeter of the triangle, respectively, and $t$ be any real number. Prove that
$$t s^{2} \leq(t+1)^{2} R^{2}-2\left(t^{2}-5 t+2\right) R r+(4-t) r^{2}$$ | Prove that let $x=s-a, y=s-b, z=s-c$, then we have
$$s=\sigma_{1}, \quad R \cdot r=\frac{\sigma_{1} \sigma_{2}-\sigma_{3}}{4 \sigma_{1}}, \quad r^{2}=\frac{\sigma_{3}}{\sigma_{1}}, \quad R^{2}=\frac{\left(\sigma_{1} \sigma_{2}-\sigma_{3}\right)^{2}}{16 \sigma_{1} \sigma_{3}}$$
Thus, (3.2.14) is equivalent to
$$L_{8}=(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,349 |
Example 9 Let $x, y, z \geq 0, x^{2}+y^{2}+z^{2}=1, g=x+y+z-x y z$.
(1) Prove that the minimum value of $g$ is 1;
(2) Prove or disprove that $g \leq \frac{8 \sqrt{3}}{9}$ (Note: The above problem was proposed by Professor Huang Yumin). | (1) Since $x^{2}+y^{2}+z^{2}=1$, we have $x+y+z \geq 3 \sqrt[3]{x y z} \geq 3 x y z$, so $g=x+y+z-x y z>0$, and $\sigma_{1}^{2}-2 \sigma_{2}=x^{2}+y^{2}+z^{2}=1$, which gives $g=\sigma_{1}\left(\sigma_{1}^{2}-2 \sigma_{2}\right)-\sigma_{3}$,
$$\begin{aligned}
g \geq 1 & \Longleftrightarrow\left(\sigma_{1}^{3}-2 \sigma_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,350 |
Example 10 Prove
$$\sum x^{3}(x-y) \geq \frac{1}{3} \sum\left(y^{2}-z^{2}\right)^{2}$$ | Prove
$$\begin{aligned}
(3.2 .19) & \Longleftrightarrow \sigma_{1}^{4}-4 \sigma_{1}^{2} \sigma_{2}+7 \sigma_{2}^{2}-3 \sigma_{1} \sigma_{3}-3 \sigma_{1} \sum x^{2} y \geq 0 \\
& \Longleftrightarrow 2 \sigma_{1}^{4}-11 \sigma_{1}^{2} \sigma_{2}+14 \sigma_{2}^{2}+3 \sigma_{1} \sigma_{3} \geq-3 \sigma_{1} \prod(x-y)
\end{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,351 |
Example 11 In $\triangle A B C$, $a, b, c$ are the lengths of the sides opposite to vertices $A, B, C$ respectively, $h_{a}, h_{b}, h_{c}$ are the lengths of the altitudes from vertices $A, B, C$ to their opposite sides, and $r_{a}, r_{b}, r_{c}$ are the radii of the excircles opposite to vertices $A, B, C$ respectivel... | Prove that by making the substitution $a=x+y, b=y+z, c=z+x, x>0, y>0, z>0$, and using the formulas
$$\begin{array}{l}
r_{a}=\frac{2 S}{b+c-a}, \quad r_{b}=\frac{2 S}{c+a-b}, \quad r_{c}=\frac{2 S}{a+b-c} \\
h_{a}=\frac{2 S}{a}, \quad h_{b}=\frac{2 S}{b}, \quad h_{c}=\frac{2 S}{c}, \quad S=\sqrt{(s-a)(s-b)(s-c) s}
\end{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,352 |
Inference 1 The necessary and sufficient condition for the symmetric cubic inequality satisfying $P_{3}(1,1,1)=0$
$$P_{3}(x, y, z)=\lambda_{0,1} \sigma_{1}^{3}+\lambda_{0,2} \sigma_{1} \sigma_{2}+\lambda_{1,1} \sigma_{3} \geq 0$$
is
$$P_{3}(1,0,0)=\lambda_{0.1} \geq 0, \quad P_{3}(1,1,0)=2\left(4 \lambda_{0,1}+\lambda... | Proof: The necessity is obviously true. Now we prove the sufficiency.
Since $P_{3}(1,1,1)=0$, according to the aforementioned method Schp, we can obtain
$$P_{3}(x, y, z)=\alpha_{0,1} f_{0.1}^{(3)}+\alpha_{0,2} f_{0,2}^{(3)}=\lambda_{0.1} f_{0.1}^{(3)}+\left(4 \lambda_{0,1}+\lambda_{0,2}\right) f_{0.2}^{(3)},$$
Therefo... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,354 |
Inference 2 Quartic Symmetric Inequality
$$P_{4}(x, y, z)=\lambda_{0,1} \sigma_{1}^{4}+\lambda_{0,2} \sigma_{1}^{2} \sigma_{2}+\lambda_{0,3} \sigma_{2}^{2}+\lambda_{1,1} \sigma_{1} \sigma_{3} \geq 0$$
Under the condition that $P_{4}(1,1,1)=0$, the necessary and sufficient condition for it to hold is
$$P_{4}(x, 1,1) \g... | Proof According to algorithm Schp, we have
$$P_{4}(x, y, z)=\alpha_{0.1} f_{0.1}^{(4)}+\alpha_{0.2} f_{0.2}^{(4)}+\alpha_{0.3} f_{0.3}^{(4)}$$
where $\alpha_{0.1}=\lambda_{0.1}, \alpha_{0,2}=5 \lambda_{0.1}+\lambda_{0,2}, \alpha_{0.3}=16 \lambda_{0,1}+4 \lambda_{0,2}+\lambda_{0,3}$, and according to (2.2.1)-(2.2.3), w... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,355 |
Inference 3 Symmetric Inequality of Degree 6
$$P_{6}(x, y, z)=\lambda_{0,3} \sigma_{1}^{2} \sigma_{2}^{2}+\lambda_{0,4} \sigma_{2}^{3}+\sigma_{3}\left(\lambda_{1,1} \sigma_{1}^{3}+\lambda_{1,2} \sigma_{1} \sigma_{2}\right)+\lambda_{2,1} \sigma_{3}^{2} \geq 0$$
Under the condition $P_{6}(1,1,1)=0$. The necessary and su... | Prove that according to the Schp method, we can obtain
$$P_{6}=\alpha_{0.3} f_{0,3}^{(6)}+\alpha_{0.4} f_{0.4}^{(6)}+\alpha_{1,1} f_{1,1}^{(6)}+\alpha_{1.2} f_{1,2}^{(6)}$$
where $\alpha_{0,3}=\lambda_{0,3}, \alpha_{0,4}=4 \lambda_{0,3}+\lambda_{0,4}, \alpha_{1,1}=4 \lambda_{0,3}+\lambda_{1,1}, \alpha_{1,2}=14 \lambda... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,356 |
Example 1 Prove
$$S \leq \frac{\sqrt{3} \sqrt[3]{a b c} \sqrt[6]{a^{2} b^{2} c^{2}-(a-b)^{2}(b-c)^{2}(c-a)^{2}}}{4}$$
where $a, b, c, S$ are the three sides and the area of an arbitrary triangle (this inequality is based on Professor Leng Gangsong's work on the Pólya inequality. | Prove that first rationalize (3.2.1) and make the substitution $a=y+z, b=x+z, c=x+y$, then the original inequality becomes
$$27 \prod(x+y)^{2}\left(\prod(x+y)^{2}-\prod(x-y)^{2}\right)-4096 \sigma_{1}^{3} \sigma_{3}^{3} \geq 0$$
Let the left side of (3.2.2) be denoted as $L_{1}(x, y, z)$. It can be verified that $L_{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,357 |
Example 2 Try to find the range of positive numbers $k, k'$ such that the following inequality holds:
$$\frac{\sum(y-z)^{2}}{4 k \sum y z} \leq \sum \frac{x}{y+z}-\frac{3}{2} \leq \frac{\sum(y-z)^{2}}{4 k^{\prime} \sum y z}(x>0, y>0, z>0)$$ | The first inequality of (3.2.4) is equivalent to
$$L_{2,1}(x, y, z)=(2 k-1) f_{0,2}^{(5)}+(k-1) f_{0,3}^{(5)}+(6 k-4) f_{1,1}^{(5)} \geq 0$$
The second inequality of (3.2.4) is equivalent to
$$L_{2,2}(x, y, z)=\left(1-2 k^{\prime}\right) f_{0,2}^{(5)}+\left(1-k^{\prime}\right) f_{0,3}^{(5)}+\left(4-6 k^{\prime}\right)... | k \geq 1, 0 \leq k^{\prime} \leq \frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,358 |
Example 3 Prove that when $k \geq 1, x>0, y>0, z>0$
$$\sum \frac{x y z}{y^{3}+z^{3}+k x y z} \leq \frac{3}{2+k}$$ | Prove that after rearranging the above equation, we obtain its equivalent form:
$$\begin{aligned}
L_{3}= & 3 f_{0.4}^{(9)}+3 f_{0.5}^{(9)}+2(k-1) f_{1,1}^{(9)}+(2 k+1) f_{1,2}^{(9)} \\
& +(2 k+1) f_{1.3}^{(9)}+2(5 k+1) f_{1,4}^{(9)}+2\left(k^{2}-k+3\right) f_{2,1}^{(9)}+\left(2 k^{2}+2 k-1\right) f_{2,2}^{(9)}
\end{ali... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,359 |
Example 4: Prove that when $k \geq 4$, we have
$$\sum \frac{x y z}{(y+z)^{3}+k x y z} \leq \frac{3}{8+k}(x>0, y>0, z>0)$$ | Prove that after rearranging the above equation, we get
$$\begin{aligned}
(3.2 .8) \Longleftrightarrow & 3 f_{0.4}^{(9)}+12 f_{0.5}^{(9)}+(-8+2 k) f_{1,1}^{(9)}+(-20+8 k) f_{1,2}^{(9)}+(1+14 k) f_{1,3}^{(9)} \\
& +(56+34 k) f_{1,4}^{(9)}+\left(-48+28 k+2 k^{2}\right) f_{2,1}^{(9)}+\left(8+8 k+5 k^{2}\right) f_{2,2}^{(9... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,360 |
Theorem 3 Let point $O$ be coplanar with $\triangle A B C$, then
$$\begin{array}{ll}
& T_{\triangle B O C} \overrightarrow{O A}+T_{\triangle A O C} \overrightarrow{O B}+T_{\triangle A O B} \overrightarrow{O C}=\mathbf{0}, \\
\text { and } & T_{\triangle B O C}+T_{\triangle A O C}+T_{\triangle A O B}=S_{\triangle A B C... | Proof: Discuss the position of point $O$ as follows:
(I) When point $O$ is inside or on the boundary of $\triangle ABC$, $\triangle ABC$ is divided into $\triangle BOC, \triangle AOC$, and $\triangle AOB$ (when $O$ is on the boundary, some of them are degenerate triangles), so we have
$$T_{\triangle BOC}=S_{\triangle B... | proof | Geometry | proof | Yes | Yes | inequalities | false | 731,367 |
Example 11 (46th IMO Problem) Let $x, y, z$ be positive real numbers. If $xyz \geqslant 1$, prove: $\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+$
$$\frac{y^{5}-y^{2}}{y^{5}+x^{2}+z^{2}}+\frac{z^{3}-z^{2}}{z^{5}+x^{2}+y^{2}} \geqslant 0$$ | Prove that since $y^{4}+z^{4}-y z\left(y^{2}+z^{2}\right)=(y-z)^{2}\left(y^{2}+y z+z^{2}\right) \geqslant 0$, then $y^{4}+z^{4} \geqslant y z\left(y^{2}+z^{2}\right)$.
Since $y^{2}+z^{2} \leqslant x y z\left(y^{2}+z^{2}\right) \leqslant x\left(y^{4}+z^{4}\right)$,
then $\frac{x^{5}}{x^{5}+y^{2}+z^{2}} \geqslant \frac{x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,369 |
5. (40th IMO Problem) $n$ is a given integer, $n \geqslant 2$, determine the smallest constant $c$, such that the inequality $\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leqslant c\left(\sum_{i=1}^{n} x_{i}\right)^{4}$ holds for all non-negative real numbers $x_{1}, x_{2}, \cdots, x_... | 5. Solution If $x_{1}, x_{2}, \cdots, x_{n}$ are all 0, $c$ can be any real number.
If $x_{1}, x_{2}, \cdots, x_{n}$ have at least one positive number. Without loss of generality, assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$. Since the inequality is homogeneous, we can set $\sum_{i=1}^{n} x_{i}=1$. ... | \frac{1}{8} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,370 |
7. Given that $a, b, c$ are non-negative real numbers, and $a+b+c=1$, prove that: $\left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2}$
$$\geqslant 2 \text {. }$$ | 7. Proof If $a, b, c$ satisfy the given conditions, then $0, a+b, c$ are also three numbers that satisfy the conditions. Let the left side of the original inequality be $f(a, b, c)$, then $f(a, b, c)-f(0, a+b, c)$
$$\begin{array}{l}
=\left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2}-\left\{\lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,371 |
8. (2003 Singapore Olympiad Question) If $x, y, z \in \mathbf{R}^{-}$, prove:
$$\frac{x y z}{(1+5 x)(4 x+3 y)(5 y+6 z)(z+18)} \leqslant \frac{1}{5120}$$ | 8. Proof Let the left side of the inequality be $f(x, y, z)$. First, fix $y$, then $f(x, y, z)$ becomes a binary function $g(x, z)=y$. $\frac{x}{(1+5 x)(4 x+3 y)} \cdot \frac{z}{(5 y+6 z)(z+18)}$,
and $\frac{x}{(1+5 x)(4 x+3 y)}=\frac{x}{20 x^{2}+(15 y+4) x+3 y}=\frac{1}{20 x+\frac{3 y}{x}+(15 y+4)}$
$$\leqslant \frac... | \frac{1}{5120} | Inequalities | proof | Yes | Yes | inequalities | false | 731,372 |
Example 2 (Adapted from the 21st question of the 2007 Zhejiang Province College Entrance Examination, Science Section) Given the sequence $\left\{a_{n}\right\}$ where the adjacent terms $a_{2 k-1}, a_{2 k}$ are the two roots of the equation $x^{2}-\left(2^{k}+3 k\right) x+3 k \cdot 2^{k}=0$, and $a_{2 k-1} \leqslant a_... | Prove that the two roots of the equation $x^{2}-\left(3 k+2^{k}\right) x+3 k \cdot 2^{k}=0$ are $x_{1}=3 k, x_{2}=2^{k}$. It is easy to find that $a_{1}=2, a_{2}=3, a_{3}=4, a_{4}=6, a_{5}=9, a_{6}=8, a_{7}=12, a_{8}=16$. When $k \geqslant 4$, it is not difficult to prove that $2^{k}>3 k$. Since $a_{2 k-1} \leqslant a_... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,374 |
Example 6 (Adapted from the 12th IMO problem) Let the sequence $\left\{a_{n}\right\}$ satisfy $1=a_{0} \leqslant a_{1} \leqslant \cdots \leqslant a_{n} \leqslant \cdots$, and the sequence $\left\{b_{n}\right\}$ is defined by: $b_{n}=\sum_{k=1}^{n}\left(1-\frac{a_{k-1}}{a_{k}}\right) \cdot \frac{1}{\sqrt{a_{k}}}, n=1,2,... | Prove that by directly decomposing the known expression,
$$\begin{array}{l}
\left(1-\frac{a_{k-1}}{a_{k}}\right) \frac{1}{\sqrt{a_{k}}}=\frac{1}{\sqrt{a_{k}}}\left(\frac{a_{k}-a_{k-1}}{a_{k}}\right)=\frac{a_{k-1}}{\sqrt{a_{k}}}\left(\frac{1}{a_{k-1}}-\frac{1}{a_{k}}\right) \\
=\frac{a_{k-1}}{\sqrt{a_{k}}}\left(\frac{1}... | 0 \leqslant b_{n} < 2 | Inequalities | proof | Yes | Yes | inequalities | false | 731,376 |
Example 9 Given the function $f(x)=\frac{1}{4^{x}+2}(x \in \mathbf{R})$, points $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right)$ are two points on the graph of the function $f(x)$, and the x-coordinate of the midpoint $P$ of the line segment $P_{1} P_{2}$ is $\frac{1}{2}$.
(1) If the general term formul... | (1) From the problem, we know: $x_{1}+x_{2}=2 \times \frac{1}{2}=1$, so,
$$\begin{aligned}
y_{1}+y_{2} & =f\left(x_{1}\right)+f\left(x_{2}\right)=\frac{1}{4^{x_{1}}+2}+\frac{1}{4^{x_{2}}+2}=\frac{4^{x_{1}}+4^{x_{2}}+4}{\left(4^{x_{1}}+2\right)\left(4^{x_{2}}+2\right)} \\
& =\frac{4^{x_{1}}+4^{x_{2}}+4}{4^{x_{1}+x_{2}}+... | a<\frac{5}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,377 |
Example 11 Given an infinite sequence of positive numbers $\left\{a_{n}\right\}$ satisfies:
(1) There exists $m \in \mathbf{R}$, such that $a_{i} \leqslant m(i=1,2, \cdots)$;
(2) For any positive integers $i, j(i \neq j)$, we have $\left|a_{i}-a_{j}\right| \geqslant \frac{1}{i+j}$. Prove: $m \geqslant 1$. | Prove that for $n \geqslant 4$, let $k_{1}, k_{2}, \cdots, k_{n}$ be a permutation of $1,2, \cdots, n$, and satisfy
$$0<a_{k_{1}}<a_{k_{2}}<\cdots<a_{k_{n}} \leqslant m$$
From condition (2), we have $a_{k_{i}}-a_{k_{i-1}} \geqslant \frac{1}{k_{i}+k_{i-1}}(i=2,3, \cdots, n)$.
Thus, for any $n \geqslant 4$, we have
$$\b... | m \geqslant 1 | Inequalities | proof | Yes | Yes | inequalities | false | 731,378 |
Example 14 (2nd Women's Olympiad Problem) The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=2, a_{n+1}=a_{n}^{2}-a_{n}+1, n=$ $1,2, \cdots$, prove:
$$1-\frac{1}{2003^{2003}}<\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{2003}}<1 .$$ | Given: $a_{n+1}-1=a_{n}\left(a_{n}-1\right)$, so
$$\frac{1}{a_{n+1}-1}=\frac{1}{a_{n}-1}-\frac{1}{a_{n}}$$
Therefore, $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{2003}}$
$$\begin{array}{l}
=\left(\frac{1}{a_{1}-1}-\frac{1}{a_{2}-1}\right)+\left(\frac{1}{a_{2}-1}-\frac{1}{a_{3}-1}\right)+\cdots+\left(\frac{1}{a... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 731,379 |
Example 15 Given a positive integer $n \geqslant 2$, let positive integers $a_{i}(i=1,2, \cdots, n)$ satisfy $a_{1}<a_{2}<\cdots<a_{n}$ and $\sum_{i=1}^{n} \frac{1}{a_{i}} \leqslant 1$. Prove: For any real number $x$, we have
$$\left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} \leqslant \frac{1}{2} \cdot \frac{... | When $x^{2} \geqslant a_{1}\left(a_{1}-1\right)$, from $\sum_{i=1}^{n} \frac{1}{a_{i}} \leqslant 1$, we can get
$$\begin{aligned}
\left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} & \leqslant\left(\sum_{i=1}^{n} \frac{1}{2 a_{i}|x|}\right)^{2}=\frac{1}{4 x^{2}}\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right)^{2} \\
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,380 |
5. Given that $f(x)$ is a linear function, $f[f(1)]=-1$, and the image of $f(x)$ symmetric to the line $x-y=0$ is $C$. If the point $\left(n, \frac{a_{n+1}}{a_{n}}\right)\left(n \in \mathbf{N}^{\cdot}\right)$ lies on the curve $C$, and $a_{1}=1, \frac{a_{n+1}}{a_{n}}-\frac{a_{n}}{a_{n-1}}=1(n \geqslant 2)$.
(1) Find th... | 5. Solution (1) Let $f(x)=k x+b(k \neq 0), \therefore f[f(1)]=k^{2}+k b+b=-1$. (1) Since the image of $f(x)$ is symmetric about the line $x-y=0$, the curve $C$ is: $f^{-1}(x)=\frac{x}{k}-\frac{b}{k}, \therefore f^{-1}(n)=\frac{n}{k}-\frac{b}{k}, f^{-1}(n-1)=\frac{n-1}{k}-\frac{b}{k}, f^{-1}(n)-f^{-1}(n-1)=\frac{1}{k}$,... | 0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,381 |
6. A point moving in the positive direction of the $O x$ axis has a horizontal coordinate of $x(t)=5(t+1)^{2}+\frac{a}{(t+1)^{5}}$, where $a$ is a positive constant. Find the minimum value of $a$ that satisfies $x(t) \geqslant 24$ for all $t \geqslant 0$. | 6. Solution Using the AM-GM inequality $x(t)=5(t+1)^{2}+\frac{a}{2(t+1)^{3}}+\frac{a}{2(t+1)^{5}} \geqslant 7 \sqrt[7]{\frac{a^{2}}{4}}$, equality holds when $(t+1)^{2}=\frac{a}{2(t+1)^{3}}$, thus we have $7 \sqrt[7]{\frac{a^{2}}{4}} \geqslant 24 \Rightarrow a \geqslant 2 \sqrt{\left(\frac{24}{7}\right)^{7}}$, on the o... | 2 \sqrt{\left(\frac{24}{7}\right)^{7}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,382 |
7. Given the sequence $a_{1}=3, a_{n+1}=\left(a_{n}-1\right)^{2}+1$, prove: $\prod_{k=1}^{n} a_{k}<2^{2^{n}}$ | 7. Solution: According to the recursive relation, using the iteration method to find the general term, and by the general term, we can prove the original formula. We have $a_{n+1}-1=\left(a_{n}-1\right)^{2}$. Using the iteration method, we get
$$a_{n-1}-1=\left(a_{n}-1\right)^{2}=\left[\left(a_{n-1}-1\right)^{2}\right]... | 2^{2^{n}}-1<2^{2^{n}} | Algebra | proof | Yes | Yes | inequalities | false | 731,383 |
8. Let $P_{1}, P_{2}, \cdots, P_{n}(n \geqslant 2)$ be any permutation of $1,2, \cdots, n$. Prove that:
$$\frac{1}{P_{1}+P_{2}}+\frac{1}{P_{2}+P_{3}}+\cdots+\frac{1}{P_{n-2}+P_{n-1}}+\frac{1}{P_{n-1}+P_{n}}>\frac{n-1}{n+2}$$ | 8. Solution Let $A=\left(\frac{1}{P_{1}+P_{2}}+\frac{1}{P_{2}+P_{3}}+\cdots+\frac{1}{P_{n-2}+P_{n-1}}+\frac{1}{P_{n-1}+P_{n}}\right)$, $B=\left(P_{1}+P_{2}\right)+\left(P_{2}+P_{3}\right)+\cdots+\left(P_{n-1}+P_{n}\right)$. Then $A \cdot B>(n-1)^{2} \cdot\left(P_{1}+P_{2} \neq P_{2}+P_{3}\right.$, so the equality does ... | \frac{n-1}{n+2} | Inequalities | proof | Yes | Yes | inequalities | false | 731,384 |
10. Let $a_{1}>a_{2}>\cdots>a_{n}>a_{n+1}$, prove that: $\frac{1}{a_{1}-a_{2}}+\frac{1}{a_{2}-a_{3}}+\cdots+\frac{1}{a_{n}-a_{n+1}}+\frac{1}{a_{n+1}-a_{1}}$
$$>0 .$$ | 10. Proof changed to: Prove $\left(a_{1}-a_{n+1}\right)\left[\frac{1}{a_{1}-a_{2}}+\frac{1}{a_{2}-a_{3}}+\cdots+\frac{1}{a_{n}-a_{n+1}}\right]>1$, write $a_{1}-a_{n+1}$ as $a_{1}-a_{n+1}=\left(a_{1}-a_{2}\right)+\left(a_{2}-a_{3}\right)+\cdots+\left(a_{n}-a_{n+1}\right)$, thus
$\left[\left(a_{1}-a_{2}\right)+\left(a_{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,385 |
11. (Mathematical Bulletin 2007 Problem 1677) Given the function $f(x)=\frac{2^{x}}{2^{x}+1}, a \leqslant b \leqslant c$, prove:
$$f(a-b)+f(b-c)+f(c-a) \leqslant \frac{3}{2} .$$ | 11. Prove (1) For any real number $m$, we have
$$f(m)+f(-m)=\frac{2^{m}}{2^{m}+1}+\frac{2^{-m}}{2^{-m}+1}=\frac{2^{m}}{2^{m}+1}+\frac{1}{2^{m}+1}=1$$
(2) For $m \geqslant 0, n \geqslant 0$, we have $2^{m-n}+2^{-m}-2^{n}-1=2^{n}\left(2^{m}-1\right)-2^{-m}\left(2^{m}-1\right)=\left(2^{n}-2^{-m}\right)\left(2^{m}-1\right)... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 731,386 |
12. Given the function $f(x)=\mathrm{e}^{x}-k x, x \in \mathbf{R}$, let the function $F(x)=f(x)+f(-x)$, prove that: $F(1) F(2) \cdots F(n)>\left(\mathrm{e}^{n+1}+2\right)^{\frac{n}{2}}\left(n \in \mathbf{N}^{*}\right)$ | 12. Solution The original inequality is transformed into: $\left(\mathrm{e}+\mathrm{e}^{-1}\right)\left(\mathrm{e}^{2}+\mathrm{e}^{-2}\right)\left(\mathrm{e}^{3}+\mathrm{e}^{-3}\right) \cdots\left(\mathrm{e}^{n}+\mathrm{e}^{-n}\right)>\left(\mathrm{e}^{n+1}+2\right)^{\frac{n}{2}}\left(n \in \mathbf{N}^{*}\right)$. Let ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,387 |
13. Given the sequence $\left\{a_{n}\right\}$, the sum of the first $n$ terms $S_{n}$ satisfies $S_{n}=2 a_{n}+(-1)^{n}, n \geqslant 1$
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Prove: For any integer $m>4$, $\frac{1}{a_{4}}+\frac{1}{a_{5}}+\cdots+\frac{1}{a_{m}}<\frac{7}{8}$. | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,388 |
14. Given that $f(n)$ is defined on the set of positive integers, and satisfies $f(1)=2, f(n+1)=(f(n))^{2}-f(n)+$ $1, n \in \mathbf{N}^{*}$, prove that for all integers $n>1$, we have $1-\frac{1}{2^{2^{2-1}}}<\sum_{i=1}^{n} \frac{1}{f(i)}<1-\frac{1}{2^{2^{2}}}$. | 14. Proof: By the given condition, it is obvious that $f(n) \geqslant 2$.
Transform $f(n+1)=(f(n))^{2}-f(n)+1$ into $f(n+1)-1=f(n)[f(n)-1]$, then $\frac{1}{f(n+1)-1}=$ $\frac{1}{f(n)-1}-\frac{1}{f(n)}$.
Thus, $\sum_{i=1}^{n} \frac{1}{f(i)}=\sum_{i=1}^{n}\left(\frac{1}{f(i)-1}-\frac{1}{f(i+1)-1}\right)=\frac{1}{f(1)-1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,389 |
15. (2005 National Training Team Training Question) The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=\frac{1}{2}$, and
$$a_{n+1}=\frac{a_{n}^{2}}{a_{n}^{2}-a_{n}+1}, n=1,2, \cdots$$
Prove: For every positive integer $n$, we have $a_{1}+a_{2}+\cdots+a_{n}<1$. | 15. Proof Since $a_{n}^{2}-a_{n}+1=\left(a_{n}-\frac{1}{2}\right)^{2}+\frac{3}{4}>0$, we have $a_{n}>0, n \in \mathbf{N}^{*}$, and
$$a_{n+1}=\frac{a_{n}^{2}}{a_{n}^{2}-a_{n}+1} \leqslant \frac{a_{n}^{2}}{2 a_{n}-a_{n}}=a_{n}$$
Therefore, $a_{n-1} \leqslant a_{n}, n \in \mathbf{N}^{+}$. Thus,
$$\begin{array}{l}
a_{n}=\... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,390 |
Example 1 (Abel's Inequality) Let $b_{1} \geqslant b_{2} \geqslant \cdots \geqslant b_{n}>0, m \leqslant \sum_{k=1}^{t} a_{k} \leqslant M, t=1,2, \cdots, n$ then we have $b_{1} m \leqslant \sum_{k=1}^{n} a_{k} b_{k} \leqslant b_{1} M$ | Prove that by Abel's summation formula, we get
$$\begin{aligned}
\sum_{k=1}^{n} a_{k} b_{k} & =b_{n}\left(\sum_{k=1}^{n} a_{k}\right)+\sum_{k=1}^{n-1}\left(\sum_{i=1}^{k} a_{i}\right)\left(b_{k}-b_{k-1}\right) \\
& \leqslant M\left[b_{n}+\sum_{k=1}^{n-1}\left(b_{k}-b_{k+1}\right)\right]=b_{1} M,
\end{aligned}$$
Simila... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,391 |
Example 3 Prove the Lagrange identity:
$$\left(\sum_{i=1}^{n} a_{i}^{2}\right) \cdot\left(\sum_{i=1}^{n} b_{i}^{2}\right)=\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{2}+\sum_{1 \leqslant i \leqslant j \leqslant n}\left(a_{i} b_{j}-a_{j} b_{i}\right)^{2},$$
and use this identity to show that the Cauchy inequality holds. | $$\begin{aligned}
& \left(\sum_{i=1}^{n} a_{i}^{2}\right) \cdot\left(\sum_{i=1}^{n} b_{i}^{2}\right)-\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{2} \\
= & \sum_{i=1}^{n} \sum_{j=1}^{n} a_{i}^{2} b_{j}^{2}-\sum_{i=1}^{n} \sum_{j=1}^{n} a_{i} b_{i} a_{j} b_{j} \\
= & \frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n}\left(a_{i}^{2} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,392 |
Example 5 Given that $a_{1}, a_{2}, a_{3}, \cdots$ is a sequence of pairwise distinct natural numbers, prove that for any positive integer $n$ we have $\sum_{k=1}^{n} \frac{a_{k}}{k^{2}}$
$$\geqslant \sum_{k=1}^{n} \frac{1}{k}$$ | Prove that by Abel's summation formula, and $\sum_{i=1}^{k} a_{i} \geqslant \sum_{i=1}^{k} i \ (1 \leqslant k \leqslant n)$, we have
$$\begin{aligned}
\sum_{k=1}^{n} \frac{a_{k}}{k^{2}} & =\frac{1}{n^{2}} \sum_{k=1}^{n} a_{k}+\sum_{k=1}^{n-1}\left(\sum_{i=1}^{k} a_{i}\right)\left[\frac{1}{k^{2}}-\frac{1}{(k+1)^{2}}\rig... | \sum_{k=1}^{n} \frac{1}{k} | Inequalities | proof | Yes | Yes | inequalities | false | 731,393 |
Example 6 Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}>0, b_{1} \geqslant a_{1}, b_{1} b_{2} \geqslant a_{1} a_{2}, \cdots, b_{1} b_{2} \cdots b_{n} \geqslant a_{1} a_{2} \cdots a_{n}$, prove that:
$$\sum_{i=1}^{n} b_{i} \geqslant \sum_{i=1}^{n} a_{i}$$ | Prove: Obviously, $b_{1}, b_{2}, \cdots, b_{n}$ are all positive numbers. Let $c_{i}=\frac{b_{i}}{a_{i}}$, then $c_{1} \geqslant 1, c_{1} c_{2} \geqslant 1, \cdots, c_{1} c_{2} \cdots c_{n} \geqslant 1$. We need to prove $\sum_{k=1}^{n} a_{k}\left(c_{k}-1\right) \geqslant 0$. From $\sum_{i=1}^{k} c_{i} \geqslant k \sqr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,394 |
Example 10 If $a_{i}>0, b_{i}>0, a_{i} b_{i}=c_{i}^{2}+d_{i}^{2}(i=1,2, \cdots, n)$, then
$$\left(\sum_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{n} b_{i}\right) \geqslant\left(\sum_{i=1}^{n} c_{i}\right)^{2}+\left(\sum_{i=1}^{n} d_{i}\right)^{2}$$
Equality holds if and only if $\frac{a_{i}}{a_{j}}=\frac{b_{i}}{b_{j}}=\f... | Prove $\left(\sum_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{n} b_{i}\right)-\left(\sum_{i=1}^{n} c_{i}\right)^{2}-\left(\sum_{i=1}^{n} d_{i}\right)^{2}$
$$\begin{array}{l}
=\sum_{i=1}^{n} \sum_{j=1}^{n} a_{i} b_{j}-\sum_{i=1}^{n} \sum_{j=1}^{n} c_{i} c_{j}-\sum_{i=1}^{n} \sum_{j=1}^{n} d_{i} d_{j} \\
=\sum_{i=1}^{n} \sum... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,395 |
1. (Chung Kai-Lai Inequality) Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}>0, \sum_{i=1}^{k} a_{i} \leqslant \sum_{i=1}^{k} b_{i}(1 \leqslant k \leqslant n)$, then we have (1) $\sum_{i=1}^{n} a_{i}^{2} \leqslant \sum_{i=1}^{n} b_{i}^{2}$; (2) $\sum_{i=1}^{n} a_{i}^{3} \leqslant \sum_{i=1}^{n} a_{i} b_{i}... | 1. Proof (1) By Abel's partial summation formula, we get
$$\sum_{i=1}^{n} a_{i}^{2}=a_{n} \sum_{i=1}^{n} a_{i}+\sum_{k=1}^{n-1}\left(\sum_{i=1}^{k} a_{i}\right)\left(a_{k}-a_{k+1}\right) \leqslant a_{n} \sum_{i=1}^{n} b_{i}+\sum_{k=1}^{n-1}\left(\sum_{i=1}^{k} b_{i}\right)\left(a_{k}-a_{k+1}\right)=\sum_{k=1}^{n} a_{k}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,396 |
2. (Chebyshev's Inequality) Let $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}, b_{1} \leqslant b_{2} \leqslant \cdots \leqslant b_{n}$, then
$$n \cdot \sum_{k=1}^{n} a_{k} b_{k} \geqslant\left(\sum_{k=1}^{n} a_{k}\right) \cdot\left(\sum_{k=1}^{n} b_{k}\right) \geqslant n \cdot \sum_{k=1}^{n} a_{k} b_{n-k+1} .... | 2. Proof: It is only necessary to prove the left inequality. For this, we extend the sequence $\left\{b_{n}\right\}$ so that $b_{n+1}=b_{1}(t=1,2, \cdots)$,
Notice that $\sum_{i=1}^{n} b_{i-1}=\sum_{i=1}^{n} b_{i}, \sum_{i=1}^{k} b_{i-t} \geqslant \sum_{i=1}^{k} b_{i}, a_{k}-a_{k-1} \leqslant 0(1 \leqslant k \leqslant... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,397 |
3. (Rearrangement Inequality) Let there be two ordered arrays: $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$ and $b_{1} \leqslant b_{2} \leqslant \cdots \leqslant b_{n}$. Prove: $a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}$ (in-order sum) $\geqslant a_{1} b_{j_{1}}+a_{2} b_{j_{2}}+\cdots+a_{n} b_{j_{n}}$ (dis... | 3. Proof Let $s_{i}=b_{1}+b_{2}+\cdots+b_{i}, s_{i}^{\prime}=b_{j 1}+b_{j 2}+\cdots+b_{j i}(i=1,2, \cdots, n)$. From the problem, it is easy to see that $s_{i} \leqslant s_{i}^{\prime}(i=1,2, \cdots, n-1), s_{n}=s^{\prime}{ }_{n}$.
Since $a_{i}-a_{i-1} \leqslant 0$, it follows that $s_{i}\left(a_{i}-a_{i-1}\right) \geq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,398 |
4. Let $a_{1}, a_{2}, \cdots$ be a sequence of positive real numbers, and for all $i, j=1,2, \cdots$ satisfy $a_{i+j} \leqslant a_{i}+a_{j}$. Prove that for any positive integer $n$, we have $a_{1}+\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n} \geqslant a_{n}$. | 4. Prove using Abel's transformation. Let $s_{i}=a_{1}+a_{2}+\cdots+a_{i}, i=1,2, \cdots, n$. Define $s_{0}=0$, then $2 s_{i}=\left(a_{1}+a_{i}\right)+\cdots+\left(a_{i}+a_{1}\right) \geqslant i a_{i-1}$. That is, $s_{i} \geqslant \frac{i}{2} \cdot a_{i+1}$.
$$\begin{array}{l}
\text { Hence } \sum_{i=1}^{n} \frac{a_{i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,399 |
6. Let $x_{1}, x_{2}, \cdots, x_{n}$ be non-negative real numbers, and let $x_{n+1}=x_{1}, a=\min \left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$. Prove that:
$$\sum_{j=1}^{n} \frac{1+x_{j}}{1+x_{j+1}} \leqslant n+\frac{1}{(1+a)^{2}} \sum_{j=1}^{n}\left(x_{j}-a\right)^{2}$$
Equality holds if and only if $x_{1}=x_{2}=\cdot... | 6. Proof Rewrite the inequality to be proved as
$$\sum_{j=1}^{n} \frac{1+x_{j}}{1+x_{j-1}} \leqslant \sum_{j=1}^{n} 1+\sum_{j=1}^{n} \frac{x_{j}-x_{j-1}}{1+a}+\sum_{j=1}^{n} \frac{\left(x_{j-1}-a\right)^{2}}{(1+a)^{2}}$$
It suffices to prove that for $j=1,2, \cdots, n$, we have $\frac{1+x_{j}}{1+x_{j-1}} \leqslant 1+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,400 |
7. Let $n>1, a_{1}, a_{2}, \cdots, a_{n}>0, b_{1}, b_{2}, \cdots, b_{n} \in \mathbf{R}$, and not all zero, $\sum_{i \neq j} a_{i} b_{j}=0$, prove:
$$\sum_{i \neq j} b_{i} b_{j}=0$$ | 7. Prove $\quad \sum_{i \neq j} a_{i} b_{j}=0 \Leftrightarrow \sum_{i=1}^{n} a_{i} \sum_{j=1}^{n} b_{j}=\sum_{i=1}^{n} a_{i} b_{j}$
and $2 \sum_{i \neq j} b_{i} b_{j}=\left(\sum_{i=1}^{n} b_{i}\right)^{2}-\sum_{i=1}^{n} b_{i}^{2}=\frac{\left(\sum a_{i} b_{j}\right)^{2}}{\left(\sum a_{i}\right)^{2}}-\sum_{i=1}^{n} b_{i... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,401 |
8. Let $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}, \cdots, b_{n}$ be positive real numbers, prove that:
$$\sum_{k=1}^{n}\left(a_{k}+b_{k}\right) \cdot \sum_{k=1}^{n} \frac{a_{k} b_{k}}{a_{k}+b_{k}} \leqslant\left(\sum_{k=1}^{n} a_{k}\right)\left(\sum_{k=1}^{n} b_{k}\right)$$
Equality holds if and only if $\frac{a_{k}... | 8. Proof
$$\text { Let } \begin{aligned}
D & =\left(\sum_{k=1}^{n} a_{k}\right)\left(\sum_{k=1}^{n} b_{k}\right)-\left(\sum_{k=1}^{n}\left(a_{k}+b_{k}\right)\right)\left(\sum_{k=1}^{n} \frac{a_{k} b_{k}}{a_{k}+b_{k}}\right) \\
& =\left(\sum_{k=1}^{n} a_{k}\right)\left(\sum_{j=1}^{n} b_{j}\right)-\left(\sum_{k=1}^{n}\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,402 |
Example 2 Positive numbers $x, y, z$ satisfy the system of equations $\left\{\begin{array}{l}x^{2}+x y+\frac{1}{3} y^{2}=25, \\ \frac{1}{3} y^{2}+z^{2}=9, \\ z^{2}+x z+x^{2}=16 .\end{array}\right.$ Find the value of $x y+2 y z+3 z x$. | Solve: Transform the original system of equations into
\[
\left\{
\begin{array}{l}
x^{2}-2 x\left(\frac{\sqrt{3}}{3} y\right) \cos 150^{\circ}+\left(\frac{\sqrt{3}}{3} y\right)^{2}=5^{2}, \\
\left(\frac{\sqrt{3}}{3} y\right)^{2}+z^{2}=3^{2}, \\
z^{2}-2 x z \cos 120^{\circ}+x^{2}=4^{2}.
\end{array}
\right.
\]
From this,... | 24 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,404 |
Example 3 Let $x, y, z$ be real numbers, prove: for any $\triangle ABC$, we have $x^{2}+y^{2}+z^{2} \geqslant 2 x y \cos C+$ $2 y z \cos A+2 z x \cos B$ | Consider the construction of the quadratic function $f(x)=x^{2}+y^{2}+z^{2}-(2 x y \cos C+2 y z \cos A+2 z x \cos B)$
$$=x^{2}-(2 y \cos C+2 z \cos B) x+y^{2}+z^{2}-2 y z \cos A \text {. }$$
Consider the discriminant:
$$\begin{array}{l}
\Delta=(2 y \cos C+2 z \cos B)^{2}-4\left(y^{2}+z^{2}-2 y z \cos A\right) \\
\text... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,405 |
Example 8 If $x, y, z \in \mathbf{R}^{+}$, and $x+y+z=1, n$ is a positive integer, prove:
$$\frac{x^{4}}{y\left(1-y^{n}\right)}+\frac{y^{4}}{z\left(1-z^{n}\right)}+\frac{z^{4}}{x\left(1-x^{n}\right)} \geqslant \frac{3^{n}}{3^{n+2}-9} .$$ | Given that $1-x^{n}>0,1-y^{n}>0,1-z^{n}>0$. Construct the vectors
$$\begin{aligned}
\boldsymbol{a} & =\left(\frac{x^{2}}{\sqrt{y\left(1-y^{n}\right)}}, \frac{y^{2}}{\sqrt{z\left(1-z^{n}\right)}}, \frac{z^{2}}{\sqrt{x\left(1-x^{n}\right)}}\right) \\
\boldsymbol{b} & =\left(\sqrt{y\left(1-y^{n}\right)}, \sqrt{z\left(1-z^... | \frac{3^{n}}{3^{n+2}-9} | Inequalities | proof | Yes | Yes | inequalities | false | 731,407 |
Example 10 Given that $a, b, x, y$ satisfy the system of equations
$$\left\{\begin{array}{l}
a x+b y=3, \\
a x^{2}+b y^{2}=7, \\
a x^{3}+b y^{3}=16, \\
a x^{4}+b y^{4}=42
\end{array}\right.$$
Find the value of $a x^{5}+b y^{5}$. | Given that the expression to be found has a high power, but if we notice that the expression to be found and the known expression share a common feature, that is, the coefficients of $x, y$ in each equation are the same, and the degree of the equation increases by one each time, we can use $x+y$ to multiply each equati... | 20 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,408 |
Example 11 (41st IMO Problem) Let $a, b, c$ be positive numbers, and $abc=1$. Prove that:
$$\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leqslant 1 .$$ | From the conditions, we have \( c = \frac{1}{ab} \) and \( \frac{1}{b} = ac \). Substituting these into the inequality and simplifying, we get
$$(a-1+ac)(a+1-ac)(ac+1-a) \leqslant a \cdot ac \cdot 1.$$
Equation (1) is cyclically symmetric with respect to \(a\), \(ac\), and \(1\). Without loss of generality, assume \(a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,409 |
6. Given $\left\{\begin{array}{l}a \sin x+b \cos x=0 \\ A \sin 2 x+B \cos 2 x=C\end{array}\right.$, where $a \neq 0$, prove: $2 a b A+\left(b^{2}-a^{2}\right) B+\left(a^{2}+b^{2}\right) C$ $=0$ | 6. Solution From the second equation, we get $2 A \sin x \cos x+2 B \cos ^{2} x=(B+C)\left(\sin ^{2} x+\cos ^{2} x\right)$
which is $2 A \tan x+2 B=(B+C)\left(\tan ^{2} x+1\right)$
Therefore, $2 A\left(-\frac{b}{a}\right)+2 B=(C+B)\left(\frac{b^{2}}{a^{2}}+1\right)$ Simplifying the above equation yields the required i... | 2 a b A+\left(b^{2}-a^{2}\right) B+\left(a^{2}+b^{2}\right) C=0 | Algebra | proof | Yes | Yes | inequalities | false | 731,410 |
3. Let real numbers $x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3}$ satisfy $x_{1}^{2}+x_{2}^{2}+x_{3}^{2} \leqslant 1$, prove:
$$\left(x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}-1\right)^{2} \geqslant\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-1\right) \cdot\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}-1\right) .$$ | 3. Prove that when $x_{1}^{2}+x_{3}^{2}+x_{3}^{2}=1$, the above equation obviously holds. When $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}<1$, construct a quadratic function:
$$\begin{aligned}
f(t) & =\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-1\right) t^{2}-2\left(x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}-1\right) t+\left(y_{1}^{2}+y_{2}^{2}+y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,411 |
7. Let $x, y, z$ be non-negative real numbers, and $x+y+z=1$, prove that: $0 \leqslant x y+y z+z x-2 x y z \leqslant$
$$\frac{7}{27}$$ | 7. By symmetry, without loss of generality, assume $x \geqslant y \geqslant z \geqslant 0$. From $(x+y)+z=2 \times \frac{1}{2}$, we know that $x+y, \frac{1}{2}, z$ form an arithmetic sequence. Hence, let $x+y=\frac{1}{2}+d, z=\frac{1}{2}-d$. From $x+y \geqslant 2 z$, we get $\frac{1}{6} \leqslant d \leqslant \frac{1}{2... | 0 \leqslant x y+y z+z x-2 x y z \leqslant \frac{7}{27} | Inequalities | proof | Yes | Yes | inequalities | false | 731,412 |
11. Let $a, b$ be positive real numbers, and $|\alpha|>a$, then for any value of $\theta$, we always have
$$\frac{\alpha \beta-\sqrt{a^{2} \beta^{2}+b^{2} \alpha^{2}-a^{2} b^{2}}}{\alpha^{2}-a^{2}} \leqslant \frac{\beta+b \sin \theta}{\alpha+a \cos \theta} \leqslant \frac{\alpha \beta+\sqrt{a^{2} \beta^{2}+b^{2} \alpha... | 11. Prove that $k=\frac{\beta+b \sin \theta}{\alpha+a \cos \theta}$ represents the slope of the line connecting point $A(\alpha, \beta)$ and point $P(-a \cos \theta,-b \sin \theta)$.
Point $P$ lies on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. Since $|\alpha|>a$, point $(\alpha, \beta)$ is to the right o... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,413 |
8. Let $a, b, c$ be positive real numbers, and satisfy $a b c=1$, prove that: $\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant$
$$\frac{3}{2} .$$ | 8. Proof Let $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$, then $x y z=1$. Thus, the problem is transformed into proving
$$\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geqslant \frac{3}{2}$$
and $[(y+z)+(z+x)+(x+y)] \cdot\left[\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y}\right] \geqslant(x+y+z)^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,414 |
Example 1 (5th IMO Problem) Prove: $\cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}=\frac{1}{2}$. | Proof: Let $A=\cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}$, and construct its complementary conjugate $B=\sin \frac{\pi}{7}-\sin \frac{2 \pi}{7}+$ $\sin \frac{3 \pi}{7}$
Then $A^{2}+B^{2}=3-4 \cos \frac{\pi}{7}+2 \cos \frac{2 \pi}{7}$,
$A^{2}-B^{2}=-\cos \frac{\pi}{7}+3 \cos \frac{2 \pi}{7}-5 \cos \fr... | \frac{1}{2} | Algebra | proof | Yes | Yes | inequalities | false | 731,417 |
Example 2 Given $a, b, c, d \in \mathbf{R}, a^{2}+b^{2}+c^{2}+d^{2} \leqslant 1$, prove: $(a+b)^{4}+(a+c)^{4}+(a+d)^{4}$
$$+(b+c)^{4}+(b+d)^{4}+(c+d)^{4} \leqslant 6 .$$ | Let the left side of the inequality be denoted as $A$, and construct the conjugate of $A$ as $B=(a-b)^{4}+(a-c)^{4}+(a-d)^{4}+$ $(b-c)^{4}+(b-d)^{4}+(c-d)^{4}$.
Thus, we have
$$\begin{aligned}
A+B & =6\left(a^{4}+b^{4}+c^{4}+d^{4}+2 a^{2} b^{2}+2 a^{2} c^{2}+2 a^{2} d^{2}+2 b^{2} c^{2}+2 b^{2} d^{2}+2 c^{2} d^{2}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,418 |
Example 3 If $a, b, c$ are positive numbers, prove that: $\frac{a^{3}}{a^{2}+a b+b^{2}}+\frac{b^{3}}{b^{2}+b c+c^{2}}+\frac{c^{3}}{c^{2}+c a+a^{2}} \geqslant$
$$\frac{a+b+c}{3}$$ | Let the left side of the inequality be denoted as $M$, and construct the conjugate expression $N=\frac{b^{3}}{a^{2}+a b+b^{2}}+\frac{c^{3}}{b^{2}+b c+c^{2}}+$ $\frac{a^{3}}{c^{2}+c a+a^{2}}$.
Then $M-N=0$, i.e., $M=N$.
Also, $M+N=(a+b) \cdot \frac{a^{2}-a b+b^{2}}{a^{2}+a b+b^{2}}+(b+c) \cdot \frac{b^{2}-b c+c^{2}}{b^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,419 |
Example 7 (26th Commonwealth of Independent States Mathematical Olympiad) Prove: For any real numbers $a>1, b>1$, the inequality
$$\frac{a^{2}}{b-1}+\frac{b^{2}}{a-1} \geqslant 8$$
holds. | Prove that let the left side of the inequality be $M$. Construct the conjugate expression
$$N=\frac{b^{2}}{b-1}+\frac{a^{2}}{a-1}$$
Then
$$\begin{aligned}
M-N & =\frac{a^{2}-b^{2}}{b-1}+\frac{b^{2}-a^{2}}{a-1} \\
& =\frac{(a+b)(a-b)^{2}}{(a-1)(b-1)} \geqslant 0
\end{aligned}$$
Hence, $M \geqslant N$.
And
$$\begin{ali... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,420 |
Example 8 (1991 Asia Pacific Mathematical Olympiad) Let $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}, \cdots, b_{n}$ be positive real numbers, and $\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} b_{k}$, prove: $\sum_{k=1}^{n} \frac{a_{k}^{2}}{a_{k}+b_{k}} \geqslant \frac{1}{2} \sum_{k=1}^{n} a_{k}$. | Let the left side of the inequality be denoted as $M$, and construct the pairing expression $N=\sum_{k=1}^{n} \frac{b_{k}^{2}}{a_{k}+b_{k}}$. Then, $M-N=\sum_{k=1}^{n}\left(a_{k}-b_{k}\right)=\sum_{k=1}^{n} a_{k}-\sum_{k=1}^{n} b_{k}=0$, hence $M=N$.
Also, $M+N=\sum_{k=1}^{n} \frac{a_{k}^{2}+b_{k}^{2}}{a_{k}+b_{k}}$. B... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,421 |
Example 9 (31st IMO Preliminary Test) Given $x \geqslant y \geqslant z>0$, prove that: $\frac{x^{2} y}{z}+\frac{y^{2} z}{x}+\frac{z^{2} x}{y} \geqslant x^{2}+y^{2}+z^{2}$. | Let the left side of the inequality be denoted as $M$, and construct the conjugate expression
$$N=\frac{x^{2} z}{y}+\frac{y^{2} x}{z}+\frac{z^{2} y}{x} .$$
By the Cauchy-Schwarz inequality, we have
$$M \cdot N \geqslant\left(x^{2}+y^{2}+z^{2}\right)^{2}$$
Also, $M-N$
$$\begin{array}{l}
=\frac{1}{x y z}\left[x^{3} y^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,422 |
Example 10 (6th Henan Province High School Mathematics Competition Question) If $a_{1}+a_{2}+\cdots+a_{n}=1, a_{i} \in \mathbf{R}_{+}(i=1,2$, $\cdots, n$ ), prove: $\frac{a_{1}^{4}}{a_{1}^{3}+a_{1}^{2} a_{2}+a_{1} a_{2}^{2}+a_{2}^{3}}+\frac{a_{2}^{4}}{a_{2}^{3}+a_{2}^{2} a_{3}+a_{2} a_{3}^{2}+a_{3}^{3}}+\cdots+\frac{a_... | Let the left side of the inequality be denoted as $M$, and construct the conjugate expression
$$N=\frac{a_{2}^{4}}{a_{1}^{3}+a_{1}^{2} a_{2}+a_{1} a_{2}^{2}+a_{2}^{3}}+\frac{a_{3}^{4}}{a_{2}^{3}+a_{2}^{2} a_{3}+a_{2} a_{3}^{2}+a_{3}^{3}}+\cdots+\frac{a_{1}^{4}}{a_{n}^{3}+a_{n}^{2} a_{1}+a_{n} a_{1}^{2}+a_{1}^{3}} .$$
... | \frac{1}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 731,423 |
2. Prove that: $\frac{1}{15}<\frac{1 \times 3 \times 5 \times \cdots \times 99}{2 \times 4 \times \cdots \times 100}<\frac{1}{10}$. | 2. Proof Let the middle part of the inequality be $M$. Construct the pairing expression
$$N=\frac{2}{3} \times \frac{4}{5} \times \frac{6}{7} \times \cdots \times \frac{98}{99} \times \frac{100}{101}$$
Notice that $\frac{n-1}{n}\frac{2}{3} \times \frac{4}{5} \times \frac{6}{7} \times \cdots \times \frac{98}{99} \times... | \frac{1}{15}<M<\frac{1}{10} | Inequalities | proof | Yes | Yes | inequalities | false | 731,424 |
5. (1990 National High School League Test Question) Given that $a, b$ are integers and $a>b>0, \sin \theta=\frac{2 a b}{a^{2}+b^{2}}$ (where $\theta \in \left(0, \frac{\pi}{2}\right)), A_{n}=\left(a^{2}+b^{2}\right)^{n} \sin n \theta$, prove that for all natural numbers $n, A_{n}$ is an integer. | 5. Prove the construction of the conjugate of $A_{n}$, $B_{n}=\left(a^{2}+b^{2}\right)^{n} \cos n \theta$. We prove a stronger conclusion: $A_{n}$ and $B_{n}$ are both integers.
（1）When $n=1$, $\sin \theta=\frac{2 a b}{a^{2}+b^{2}}, \cos \theta=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$. Thus, $A_{1}=2 a b, B_{1}=a^{2}-b^{2}$. B... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,426 |
8. Given that $a, b$ are positive real numbers, and $\frac{1}{a}+\frac{1}{b}=1$, prove: for every natural number $n$, we have $(a+b)^{n}-a^{n}-$ $b^{n} \geqslant 2^{2 n}-2^{n+1}$ | 8. Proof Let the left side of the inequality be $A$, transforming we get $A=C_{n}^{1} a^{n-1} b+C_{n}^{2} a^{n-2} b^{2}+\cdots+C_{n}^{-1} a b^{n-1}$.
Writing it in reverse, we have $A=C_{n}^{n-1} a^{n-1} b+C_{n}^{n-2} a^{n-2} b^{2}+\cdots+C_{n}^{1} a b^{n-1}$.
Adding the two equations, we get $2 A=C_{n}^{1}\left(a^{n-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,427 |
9. (42nd IMO Problem) For all positive real numbers $a, b, c$, prove that: $\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 a c}}+$
$$\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant 1 .$$ | 9. Prove that by the AM-GM inequality, we have
$$\begin{array}{l}
\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{\sqrt{a^{2}+8 b c}}{9 a} \geqslant \frac{2}{3}, \frac{b}{\sqrt{b^{2}+8 c a}}+\frac{\sqrt{b^{2}+8 c a}}{9 b} \geqslant \frac{2}{3} . \\
\frac{c}{\sqrt{c^{2}+8 a b}}+\frac{\sqrt{c^{2}+8 a b}}{9 c} \geqslant \frac{2}{3} . ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,428 |
10. (2005 National Training Team Test) Let $a, b, c \geqslant 0, a b+b c+c a=\frac{1}{3}$, prove:
$$\frac{1}{a^{2}-b c+1}+\frac{1}{b^{2}-c a+1}+\frac{1}{c^{2}-a b+1} \leqslant 3$$ | 10. Proof: From the given conditions, we have: \(a^{2}-b c+1 \geqslant-\frac{1}{3}+1>0\).
Similarly, \(b^{2}-c a+1>0\), \(c^{2}-a b+1>0\).
Additionally, \(a+b+c>0\). Let
\[
\begin{aligned}
M & =\frac{a}{a^{2}-b c+1}+\frac{b}{b^{2}-c a+1}+\frac{c}{c^{2}-a b+1} \\
N & =\frac{1}{a^{2}-b c+1}+\frac{1}{b^{2}-c a+1}+\frac{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,429 |
Example $\mathbf{1 1}$ (2006 Korean Olympiad Problem) For three distinct real numbers $a_{1}, a_{2}, a_{3}$, define three real numbers $b_{1}, b_{2}, b_{3}$ as follows: $b_{j}=\left(1+\frac{a_{j} a_{i}}{a_{j}-a_{i}}\right)\left(1+\frac{a_{j} a_{k}}{a_{j}-a_{k}}\right)$, where $\{i, j, k\}=\{1,2,3\}$. Prove that $1+\lef... | Proof: Let \( A = \frac{a_1 a_2}{a_1 - a_2}, B = \frac{a_1 a_3}{a_1 - a_3}, C = \frac{a_2 a_3}{a_2 - a_3} \), then
\[
\begin{aligned}
& a_1 b_1 + a_2 b_2 + a_3 b_3 = a_1 (1 + A)(1 + B) + a_2 (1 - A)(1 + C) + a_3 (1 - B)(1 - C) \\
= & a_1 + a_2 + a_3 + (a_1 - a_2) A + (a_1 - a_3) B + (a_2 - a_3) C + a_1 A B - a_2 A C + ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,430 |
Example 14 (1998 Canadian Olympiad Problem) Let $x_{1}, x_{2}, \cdots, x_{n+1}$ be positive real numbers, satisfying the condition $\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{n+1}}=1$. Prove that: $x_{1} x_{2} \cdots x_{n+1} \geqslant n^{n-1}$. | Prove that for $y_{i}=\frac{1}{1+x_{i}}, i=1,2, \cdots, n+1$, then $\sum_{i=1}^{n+1} y_{i}=1$, let $s_{i}=\sum_{1 \leqslant j \leqslant n+1 . j \neq i} y_{j}, p_{i}=$ $\prod_{1 \leqslant j \leqslant n+1, j \neq i} y_{j}$
By the arithmetic-geometric inequality, we have $\frac{1-y_{i}}{y_{i}}=\frac{s_{i}}{y_{i}} \geqsla... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,433 |
Example 15 Given $\theta_{1}, \theta_{2}, \theta_{3}, \theta_{4} \in\left(0, \frac{\pi}{2}\right)$ satisfy $\theta_{1}+\theta_{2}+\theta_{3}+\theta_{4}=\pi$, prove:
$$\frac{\sqrt{2} \sin \theta_{1}-1}{\cos \theta_{1}}+\frac{\sqrt{2} \sin \theta_{2}-1}{\cos \theta_{2}}+\frac{\sqrt{2} \sin \theta_{3}-1}{\cos \theta_{3}}+... | Let $a=\tan \theta_{1}, b=\tan \theta_{2}, c=\tan \theta_{3}, d=\tan \theta_{4}$, then $a, b, c, d>0$,
and from $\tan \left(\theta_{1}+\theta_{2}\right)+\tan \left(\theta_{3}+\theta_{4}\right)=0$ we know $\frac{a+b}{1-a b}+\frac{c+d}{1-c d}=0$.
Thus, $(a+b)(1-c d)+(c+d)(1-a b)=0 \Leftrightarrow a+b+c+d=a b c+b c d+c d ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,434 |
3. (1990 Vietnam Olympiad) Let $a, b, c \in \mathbf{R}^{+}$, and $a b c + a + c = b$, try to determine the maximum value of $p = \frac{2}{a^{2}+1} - \frac{2}{b^{2}+1} + \frac{3}{c^{2}+1}$. | 3. Solution Let $a=\tan \alpha, b=\tan \beta, c=\tan \gamma, \alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)$, then from $a b c+a+c=b$ we get $b=\frac{a+c}{1-a c}$, i.e., $\tan \beta=\tan (\alpha+\gamma)$, hence $\beta=\alpha+\gamma$.
Therefore, $p=\frac{2}{a^{2}+1}-\frac{2}{b^{2}+1}+\frac{3}{c^{2}+1}=2 \cos ^{... | \frac{10}{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,437 |
6. (33rd IMO Preliminary Problem) If $x, y, z>1$, and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$. Prove: $\sqrt{x+y+z}$ $\geqslant \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$ | 6. Proof Let $\alpha, \beta, \gamma$ all be acute angles, then $\frac{1}{\cos ^{2} \alpha}>1, \frac{1}{\cos ^{2} \beta}>1, \frac{1}{\cos ^{2} \gamma}>1.$
Make the trigonometric substitution $x=\frac{1}{\cos ^{2} \alpha}, y=\frac{1}{\cos ^{2} \beta}, z=\frac{1}{\cos ^{2} \gamma}$.
Then the condition simplifies to $\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,438 |
7. (2006 Turkey National Team Selection Test) Given positive numbers $x, y, z$ satisfying $x y+y z+z x=1$. Prove:
$$\frac{27}{4}(x+y)(y+z)(z+x) \geqslant(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^{2} \geqslant 6 \sqrt{3} .$$ | 7. Solution Let $\tan A=x, \tan B=y, \tan C=z$, where $\angle A, \angle B, \angle C \in \left(0, \frac{\pi}{2}\right)$, then
$$\begin{aligned}
& \tan A \cdot \tan B+\tan B \cdot \tan C+\tan C \cdot \tan A=1 \\
\Leftrightarrow & 1-\tan A \cdot \tan B=\tan C(\tan A+\tan B) \Leftrightarrow \cot C=\frac{\tan A+\tan B}{1-\t... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,439 |
10. Given positive numbers $m_{i} \in \mathbf{R}^{+}(i=1,2, \cdots, n), p \geqslant 2$ and $p \in \mathbf{N}$, and satisfying $\frac{1}{1+m_{1}^{P}}+\frac{1}{1+m_{2}^{P}}+\cdots$ $+\frac{1}{1+m_{n}^{p}}=1$. Prove: $m_{1} m_{2} \cdots m_{n} \geqslant(n-1)^{\frac{n}{p}}$. | 10. Proof: Let $m_{i}^{\rho}=\tan ^{2} \alpha_{i}, a_{i} \in\left(0, \frac{\pi}{2}\right)$.
From the given condition, we have $\cos ^{2} \alpha_{1}+\cos ^{2} \alpha_{2}+\cdots+\cos ^{2} \alpha_{n}=1$,
thus $\cos ^{2} \alpha_{1}+\cos ^{2} \alpha_{2}+\cdots+\cos ^{2} \alpha_{n-1}=1-\cos ^{2} \alpha_{n}=\sin ^{2} \alpha_... | m_{1} m_{2} \cdots m_{n} \geqslant(n-1)^{\frac{n}{p}} | Inequalities | proof | Yes | Yes | inequalities | false | 731,440 |
11. (2002 National Competition Question) Real numbers $a, b, c$ and a positive number $\lambda$ make $f(x)=x^{3}+a x^{2}+b x+c$ have three real roots $x_{1}, x_{2}, x_{3}$, and satisfy
(1) $x_{2}-x_{1}=\lambda$; (2) $x_{3}>\frac{1}{2}\left(x_{1}+x_{2}\right)$.
Find the maximum value of $\frac{2 a^{3}+27 c-9 a b}{\lamb... | 11. Solution: Let $x_{3}=\frac{1}{2}\left(x_{1}+x_{2}\right)+t(t>0)$.
By Vieta's formulas, we have $\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}=\frac{-2\left(x_{1}+x_{2}+x_{3}\right)^{3}+27 x_{1} x_{2} x_{3}+9\left(x_{1}+x_{2}+x_{3}\right)\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}\right)}{\lambda^{3}}$
$$=\frac{\left(x_{1}+... | \frac{3 \sqrt{3}}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,441 |
Example 8 If $a, b, c \in [0,1]$, prove:
$$\frac{a}{1+b+c}+\frac{b}{1+a+c}+\frac{c}{1+a+b}+(1-a)(1-b)(1-c) \leqslant 1$$ | $$\begin{aligned}
& \text{Left side }=\frac{a}{1+b+c}+\frac{b}{1+a+c}+\frac{c}{1+a+b}+\frac{1}{1+a+b}-\frac{1}{1+a+b} \\
& +(1-a)(1-b)(1-c) \\
\leqslant & \frac{1}{1+a+b}+\frac{a}{1+b+a}+\frac{b}{1+a+b}+\frac{c}{1+a+b}-\frac{1}{1+a+b} \\
& +(1-a)(1-b)(1-c) \\
= & \frac{1+a+b}{1+a+b}+\frac{c-1}{1+a+b}+(1-a)(1-b)(1-c) \\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,442 |
14. (34th IMO Preliminary Problem) For all positive real numbers $a, b, c, d$, prove: $\frac{a}{b+2 c+3 d}+$
$$\frac{b}{c+2 d+3 a}+\frac{c}{d+2 a+3 b}+\frac{d}{a+2 b+3 c} \geqslant \frac{2}{3} .$$ | 14. Prove that by making the following substitution in the denominator of the fraction: $\left\{\begin{array}{l}x=b+2 c+3 d, \\ y=c+2 d+3 a, \\ z=d+2 a+3 b, \\ w=a+2 b+3 c .\end{array}\right.$
Considering $a, b, c, d$ as variables, solving the system of equations yields:
$$\left\{\begin{array}{l}
a=-\frac{5}{24} x+\fr... | \frac{2}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 731,443 |
Example 2 (2004 China Mathematical Olympiad) Given a positive integer $n \geqslant 2$, let positive integers $a_{i}(i=1,2, \cdots, n)$ satisfy: $a_{1}<a_{2}<\cdots<a_{n}$ and $\sum_{i=1}^{n} \frac{1}{a_{i}} \leqslant 1$. Prove that for any real number $x$, we have $\left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^... | Given the known condition $\sum_{i=1}^{n} \frac{1}{a_{i}} \leqslant 1$ and the structure of the inequality to be proven, it is not difficult to think of using the Cauchy inequality to first perform a relaxation, obtaining $\left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} \leqslant \sum_{i=1}^{n} \frac{a_{i}}{\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,444 |
Example 3 Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1$ and $a_{n+1}=\frac{1+2 a_{n}}{1+a_{n}}\left(n \in \mathbf{N}^{*}\right)$, prove: $a_{n}<\frac{1+\sqrt{5}}{2}$. | Since the right side of the inequality is a constant, it is difficult to directly apply mathematical induction to achieve the transition from $k$ to $k+1$. Therefore, we can first prove a stronger inequality $0 < a_n < \frac{1+\sqrt{5}}{2}$. On one hand, $a_{k+1} = \frac{1 + 2a_k}{1 + a_k} > 0$; on the other hand, $a_{... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,445 |
7. Given $x_{i} \in \mathbf{R}(i=1,2, \cdots, n)$, satisfying $\sum_{i=1}^{n}\left|x_{i}\right|=1, \sum_{i=1}^{n} x_{i}=0$. Prove: $\left|\sum_{i=1}^{n} \frac{x_{i}}{i}\right| \leqslant$
$$\frac{1}{2}-\frac{1}{2 n}$$ | 7. Proof: First, strengthen the proposition to the following form: For $n(n \geqslant 2)$ real numbers $x_{1}, x_{2}, \cdots, x_{n}$, satisfying $\sum_{i=1}^{n}\left|x_{i}\right| \leqslant 1$, $\sum_{i=1}^{n} x_{i}=0$, then $\left|\sum_{i=1}^{n} \frac{x_{i}}{i}\right| \leqslant \frac{1}{2}-\frac{1}{2 n}$. Prove this us... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,449 |
10. (2004 Australian Olympiad Problem) The sequence of non-negative integers $\left\{x_{n}\right\}$ is defined as: $x_{1}$ is a non-negative integer less than 204, and $x_{n+1}=\left(\frac{n}{2004}+\frac{1}{n}\right) x_{n}^{2}-\frac{n^{3}}{2004}+1, n>0$, prove: the sequence $\left\{x_{n}\right\}$ must contain infinitel... | 10. Proof: It can be calculated that $x_{2}=\frac{2005 x_{1}^{2}+2003}{2004}=x_{1}^{2}+1+\frac{x_{1}^{2}-1}{2004}$, since all terms in the sequence are integers, therefore, 2004 $\mid x_{1}^{2}-1$. Noting that $2004=2^{2} \times 3 \times 167, 4 \mid x_{1}^{2}-1$, hence $x_{1}^{2}$ must be odd. Let $x_{1}=2 b+1$, then $... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 731,450 |
Example 1 Given positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$, satisfying $x_{1} x_{2} \cdots x_{n}=1$, prove:
$$\frac{1}{n-1+x_{1}}+\frac{1}{n-1+x_{2}}+\cdots+\frac{1}{n-1+x_{n}} \leqslant 1 .$$ | Prove by starting from a special case, then studying the general case, and solving the problem through step-by-step adjustments.
When \( x_{1} = x_{2} = \cdots = x_{n} = 1 \), the inequality holds.
When \( x_{1}, x_{2}, \cdots, x_{n} \) are not all 1, there must be one in \( (0,1) \) and one in \( (1,+\infty) \). By sy... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,451 |
Example 2 (2006 National High School Mathematics Competition Problem) Express 2006 as the sum of 5 positive integers $x_{1}, x_{2}, x_{3}, x_{4}$, $x_{5}$. Let $S=\sum_{1 \leqslant i<j \leqslant 5} x_{i} x_{j}$. Questions:
(1) For what values of $x_{1}, x_{2}, x_{3}, x_{4}$, $x_{5}$ does $S$ attain its maximum value;
(... | (1) First, the set of such $S$ values is bounded, so there must be a maximum and a minimum value.
If $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=2006$, and $S=\sum_{1 \leqslant i x_{1} x_{2}$.
Rewrite $S$ as $S=\sum_{1 \leqslant i0$, which contradicts the assumption that $S$ reaches its maximum value at $x_{1}, x_{2}, x_{3}, x_{4}... | x_{1}=402, x_{2}=x_{3}=x_{4}=x_{5}=401 \text{ for maximum } S; x_{1}=x_{2}=x_{3}=402, x_{4}=x_{5}=400 \text{ for minimum } S | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,452 |
For example, $10 a_{i}, b_{i}, c_{i}>0(i=1,2, \cdots, n)$. Prove:
$$\begin{array}{l}
\quad \sqrt[n]{\left(a_{1}+b_{1}+c_{1}\right)\left(a_{2}+b_{2}+c_{2}\right) \cdots\left(a_{n}+b_{n}+c_{n}\right)} \geqslant \sqrt[n]{a_{1} a_{2} \cdots a_{n}}+\sqrt[n]{b_{1} b_{2} \cdots b_{n}}+ \\
\sqrt[n]{c_{1} c_{2} \cdots c_{n}} .
... | Prove that this problem is not easy to tackle directly. If we transform one side of the inequality to 1, and then observe the characteristics of the new inequality structure, it is easier to find the proof idea.
The original inequality is equivalent to
$$\begin{array}{l}
\sqrt[n]{\frac{a_{1} a_{2} \cdots a_{n}}{\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,454 |
Example 5 (2003 China National Training Team Training Question) Let $x, y, z \geqslant 0, x y+y z+z x=1$. Prove: $f(x, y, z)=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x} \geqslant \frac{5}{2}$. | Assume without loss of generality that $x \geqslant y \geqslant z$, and let $u=\frac{x+y}{2} \geqslant z$. By the given condition, we have $u^{2} \geqslant \frac{1}{3}$. When $u \geqslant 1$, $f \geqslant \frac{1}{x+y}+x+y=2 u+\frac{1}{2 u} \geqslant \frac{5}{2}$; when $u>0$, thus, it suffices to prove that $f(x, y, z)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,455 |
Example 8 (40th IMO National Team Selection Test) For non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying the condition $\sum_{i=1}^{n} x_{i}=1$, find the maximum value of $\sum_{i=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)$. | Prove (1) for $x, y>0$,
$$\begin{array}{l}
(x+y)^{4}-(x+y)^{5}+0^{4}-0^{5}-\left(x^{4}-x^{5}+y^{4}-y^{5}\right) \\
=x y\left(4 x^{2}+6 x y+4 y^{2}\right)-x y\left(5 x^{3}+10 x y^{2}+10 x^{2} y+5 y^{3}\right) \\
=\frac{1}{2} x y\left(7 x^{2}+12 x y+7 y^{2}+x^{2}+y^{2}\right)-5 x y\left(x^{3}+2 x y^{2}+2 x^{2} y+y^{3}\ri... | \frac{1}{12} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,457 |
Example 9 Given $a_{1}, a_{2}, \cdots, a_{n}$ are $n(n \geqslant 3)$ positive integers, and $a_{1}+a_{2}+\cdots+a_{n}=m, m(m \geqslant n)$ is a constant. Find the maximum value of $M=\sum_{1 \leqslant i \leqslant j<k \leqslant n} a_{i} a_{j} a_{k}$. | Proof: First, we prove that when $M$ attains its maximum value $M_{1}$, for any $i, j(1 \leqslant i < j \leqslant n)$, we have $a_{i} \leqslant a_{j}+1$. Otherwise, if there exist $i, j$ such that $a_{i} \geqslant a_{j}+2$, then by setting $a_{i}^{\prime}=a_{i}-1$ and $a_{j}^{\prime}=a_{j}+1$, we get a new set of posit... | proof | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 731,458 |
Example 10 (1991 National Training Test) Let $x_{1}, x_{2}, x_{3}, x_{4}$ be positive real numbers and $x_{1}+x_{2}+x_{3}+x_{4} = \pi$. Find the minimum value of the expression $\left(2 \sin ^{2} x_{1}+\frac{1}{\sin ^{2} x_{1}}\right)\left(2 \sin ^{2} x_{2}+\frac{1}{\sin ^{2} x_{2}}\right)\left(2 \sin ^{2} x_{3}+\frac{... | Let $x_{1}+x_{2}$ be a constant. From $\sin x_{1} \cdot \sin x_{2}=\frac{1}{2}\left[\cos \left(x_{1}-x_{2}\right)-\cos \left(x_{1}+x_{2}\right)\right]$, we know that the value of $\sin x_{1} \cdot \sin x_{2}$ increases as $\left|x_{1}-x_{2}\right|$ decreases. Denote the expression under discussion as $f\left(x_{1}, x_{... | 81 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,459 |
Assume $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers. If the sum of $n$ non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ is 1, then
$$a_{1} x_{1}+a_{2} x_{3}+\cdots+a_{n} x_{n} \geq a_{1}^{x_{1}} a_{2}^{x_{2}} \cdots a_{n}^{x_{n}}$$
Proof: The proof of this inequality is similar to the classic AM-GM i... | For the case $n=2$, we must provide a detailed proof (because real number exponents appear in the inequality). We first prove that if $x, y \geq 0, x+y=1$ and $a, b>0$, then
$$a x+b y \geq a^{x} b^{y}$$
The simplest way to prove this inequality is to consider the case where $x, y$ are rational numbers. For real number... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,461 |
Example 1.1.9 Let $x, y, z$ be positive numbers, and satisfy $x^{5}+y^{5}+z^{5}=3$, prove: $\frac{x^{4}}{y^{3}}+\frac{y^{4}}{z^{3}}+\frac{z^{4}}{x^{3}} \geq 3$ | Prove: Notice that
$$\left(x^{5}+y^{5}+z^{5}\right)^{2}=x^{10}+2 x^{5} y^{5}+y^{10}+2 y^{5} z^{5}+z^{10}+2 z^{5} x^{5}=9$$
From this form, we use the AM-GM inequality, and get
$$10 \cdot \frac{x^{4}}{y^{3}}+6 x^{5} y^{5}+3 x^{10}=\underbrace{\frac{x^{4}}{y^{3}}+\cdots+\frac{x^{4}}{y^{3}}}_{10}+\underbrace{x^{5} y^{5}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,462 |
Example 4.2.9 Let real numbers $x_{1}, x_{2}, \cdots, x_{n} \in[-1,1]$, and satisfy $x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=0$, find the maximum value of the expression $x_{1}+x_{2}+\cdots+x_{n}$. (Tran Nam Dung) | Let's denote $a_{i}=x_{i}^{3} \quad i \in\{1,2, \cdots, n\}$, then $a_{1}+a_{2}+\cdots+a_{n}=0$. Note that the function $f(x)=\sqrt[3]{x}$ is concave when $x \geq 0$ and convex when $x \leq 0$. By Lemma 5, it is easy to obtain
$$f(x)+f(y) \leq\left\{\begin{array}{lr}
f(-1)+f(x+y+1) & x, y \in[-1,0] \\
2 f\left(\frac{x+... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,464 |
Theorem 8 (Abel's Formula) Suppose $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ and $\left(y_{1}, y_{2}, \cdots, y_{n}\right)$ are two sequences of real numbers.
$$\begin{array}{l}
c_{k}=y_{1}+y_{2}+\cdots+y_{k}(k=1,2, \cdots, n) \text {, then } \\
x_{1} y_{1}+x_{2} y_{2}+\cdots+x_{n} y_{n}=\left(x_{1}-x_{2}\right) c_{1}... | Proof: We of course have
$$\begin{array}{l}
\left(x_{1}-x_{2}\right) c_{1}+\left(x_{2}-x_{3}\right) c_{2}+\cdots+\left(x_{n-1}-x_{n}\right) c_{n-1}+x_{n} c_{n} \\
=c_{1} x_{1}+\left(c_{2}-c_{1}\right) x_{2}+\cdots+\left(c_{n}-c_{n-1}\right) x_{n}=x_{1} y_{1}+x_{2} y_{2}+\cdots+x_{n} y_{n}
\end{array}$$ | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,465 |
Example 5.1.1 (Abel's Inequality) Let $x_{1}, x_{2}, \cdots, x_{n}$ and $y_{1} \geq y_{2} \geq \cdots \geq y_{n} \geq 0$ be real numbers. For $k \in\{1,2, \cdots, n\}$, we denote $S_{k}=\sum_{i=1}^{n} x_{i}$, and assume $M=\max \left\{S_{1}, S_{2}, \cdots, S_{n}\right\}$ and $m=\min \left\{S_{1}, S_{2}, \cdots, S_{n}\r... | Proof: Since the proofs of the two parts of the inequality are similar, we only prove the left half of the inequality. Let \( y_{n+1}=0 \), by Abel's formula
$$\sum_{i=1}^{n} x_{i} y_{i}=\sum_{i=1}^{n}\left(y_{i}-y_{i+1}\right) S_{i} \geq \sum_{i=1}^{n} m\left(y_{i}-y_{i+1}\right)=m y_{1}$$
Abel's formula is often use... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,466 |
Example 5.1.2 Let $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1} \geq b_{2} \geq \cdots \geq b_{n} \geq 0$ be sequences of positive numbers, and satisfy $a_{1} a_{2} \cdots a_{k} \geq b_{1} b_{2} \cdots b_{k} \forall k \in\{1,2, \cdots, n\}$, prove the following inequality
$$a_{1}+a_{2}+\cdots+a_{n} \geq b_{1}+b_{2}+\cdots+b... | Proof: By Abel's formula, we have
$$\begin{array}{l}
\sum_{i=1}^{n} a_{i}-\sum_{i=1}^{n} b_{i}=\sum_{i=1}^{n} b_{i}\left(\frac{a_{i}}{b_{i}}-1\right)=\left(b_{1}-b_{2}\right)\left(\frac{a_{1}}{b_{1}}-1\right)+\left(b_{2}-b_{3}\right)\left(\frac{a_{1}}{b_{1}}+\frac{a_{2}}{b_{2}}-2\right)+\cdots \\
+\left(b_{n-1}-b_{n}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,467 |
Example 5.1.3 Let $x_{1}, x_{2}, \cdots, x_{n}$ be a sequence of positive real numbers, and satisfy $x_{1}+x_{2}+\cdots+x_{k} \geq \sqrt{k} \quad \forall k \in\{1,2, \cdots, n\}$. Prove the following inequality:
$$x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2} \geq \frac{1}{4}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\ri... | Proof: Without loss of generality, assume $x_{1} \geq x_{2} \geq \cdots \geq x_{n}$. For $k \in\{1,2, \cdots, n\}$, let $b_{k}=\frac{1}{\sqrt{k}}$, we will first prove
$$2 \sum_{i=1}^{n} x_{i}^{2} \geq \sum_{i=1}^{n} x_{i} b_{i}$$
and
$$2 \sum_{i=1}^{n} x_{i} b_{i} \geq \sum_{i=1}^{n} b_{i}^{2}$$
By Abel's formula, w... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,468 |
Example 5.1.4 Let $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ be real numbers, and satisfy
$$\begin{array}{l}
a_{1} \geq \frac{a_{1}+a_{2}}{2} \geq \cdots \geq \frac{a_{1}+a_{2}+\cdots+a_{n}}{n} \\
b_{1} \geq \frac{b_{1}+b_{2}}{2} \geq \cdots \geq \frac{b_{1}+b_{2}+\cdots+b_{n}}{n}
\end{array}$$
Pr... | Prove: For $k \in\{1,2, \cdots, n\}$, we denote $S_{k}=a_{1}+a_{2}+\cdots+a_{k}, b_{n+1}=0$. By Abel's formula, we have
$$\sum_{i=1}^{n} a_{i} b_{i}=\sum_{i=1}^{n}\left(b_{i}-b_{i+1}\right) S_{i}=\sum_{i=1}^{n} i\left(b_{i}-b_{i+1}\right)\left(\frac{S_{i}}{i}\right)$$
Again, by Abel's formula, we have
$$\begin{array}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,469 |
Example 5.1.5 Let $x_{1}, x_{2}, \cdots, x_{n}$ be real numbers, satisfying $x_{1} \geq x_{2} \geq \cdots \geq x_{n} \geq x_{n+1}=0$, prove the following inequality $\sqrt{x_{1}+x_{2}+\cdots+x_{n}} \leq \sum_{i=1}^{n} \sqrt{i}\left(\sqrt{x_{i}}-\sqrt{x_{i+1}}\right)$
(Romania MO and Singapore MO) | Prove: Let $c_{i}=\sqrt{i}-\sqrt{i-1}, a_{i}=\sqrt{x_{i}}$, then the inequality becomes
$$\left(a_{1} c_{1}+a_{2} c_{2}+\cdots+a_{n} c_{n}\right)^{2} \geq a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}$$
Assume $b_{1}, b_{2}, \cdots, b_{n}$ are positive numbers, and satisfy $b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}=1$ and $\left(\s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,470 |
Example 5.1.6 Let $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1} \leq b_{2} \leq \cdots \leq b_{n}$ be real numbers, and satisfy
$$a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2} \leq b_{1}^{2}+b_{2}^{2}+\cdots+b_{k}^{2}, k=1,2, \cdots, n \text{, prove: } a_{1}+a_{2}+\cdots+a_{n} \leq b_{1}+b_{2}+\cdots+b_{n}$$ | We prove this problem by induction. It is obvious when $n=1$. Suppose the statement is true for $n$, we will prove it is also true for $n+1$. In fact, by the Cauchy-Schwarz inequality, we have
$$\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n+1}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+\cdots+b_{n+1}^{2}\right) \geq\left(a_{1} b_{1}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,471 |
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