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Example 18 Let the quadratic function $f(x)=a x^{2}+b x+c(a \neq 0)$ have values whose absolute values do not exceed 1 on the interval $[0,1]$, find the maximum value of $|a|+|b|+|c|$. | Solve: From
we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
f(0)=c \\
f(1)=a+b+c \\
f\left(\frac{1}{2}\right)=\frac{1}{4} a+\frac{1}{2} b+c
\end{array}\right. \\
\left\{\begin{array}{l}
a=2 f(0)-4 f\left(\frac{1}{2}\right)+2 f(1) \\
b=-3 f(0)+4 f\left(\frac{1}{2}\right)-f(1) \\
c=f(0)
\end{array}\right. \\
\begin{a... | 17 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,287 |
Example 35 Let $x, y, z, w \in \mathbf{R}^{+}$, and $x+y+z+w=1$, then
$$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+w}}+\frac{w}{\sqrt{w+x}}<\frac{3}{2}$$ | Given $x+y+z+w=1$, we know that one of $x+z$ and $y+w$ must be no greater than $\frac{1}{2}$. If $y+\frac{1}{2}$, according to formula (52) in Example 34, we have
$$\begin{array}{l}
\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+w}}+\frac{w}{\sqrt{w+x}}= \\
\left(\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,289 |
Example 36 (Self-created problem, 2006.12.2) Let \(a, b, c \in \mathbf{R}^{+}\), and \(abc=1\), then
$$(b+c)(c+a)(a+b)(a+b+c) \geqslant 3\left(1+a^{2}\right)\left(1+b^{2}\right)\left(1+c^{2}\right)$$
Equality in (54) holds if and only if \(a=b=c=1\). | Expand both sides of equation (54), and let $s_{1}=a+b+c, s_{2}=b c+c a+a b$, noting that $a b c=1$, we can obtain
$$s_{1}^{2} s_{2}-3 s_{1}^{2}-3 s_{2}^{2}+5 s_{1}+6 s_{2}-6 \geqslant 0$$
Thus, to prove equation (54), it is sufficient to prove equation (1). Below, we will prove equation (1) in two cases.
i) If $s_{1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,290 |
Example 38 (Self-created problem, 2006. 10.03) Let $x, y, z \in \mathbf{R}^{-}$, and $x^{2}+y^{2}+z^{2} \leqslant 3$, then
$$2+x y z \geqslant x^{2} y+y^{2} z+z^{2} x$$
Equality in (59) holds if and only if $x=y=z=1$, or $x=0, y=\sqrt{2} z=\sqrt{2}$, or $y=0, z=\sqrt{2} x=\sqrt{2}$, or $z=0, x=\sqrt{2} y=\sqrt{2}$. | To prove that when $x \geqslant y \geqslant z$, we have
$$x^{2} y+y^{2} z+z^{2} x-\left(x y^{2}+y z^{2}+z x^{2}\right)=(x-y)(y-z)(x-z) \geqslant 0$$
Therefore, to prove equation (59), it is sufficient to prove it under the condition $x \geqslant y \geqslant z$.
Furthermore, because
$$\begin{aligned}
2+x y z-x^{2} y-y^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,292 |
Example 39 (Self-created, 2007.09.09) Let $a, b, c \in \mathbf{R}^{-}$, and $a+b+c=2$, then
$$\sum b^{3} c^{3}+\frac{537}{64}(a b c)^{3} \leqslant 1$$
Equality in (60) holds if and only if one of $a, b, c$ is zero and the other two are equal to 1. | Assume $c$ is the smallest, then $0 \leqslant c \leqslant \frac{2}{3}$, then
$$\begin{array}{l}
1-\sum b^{3} c^{3}-\frac{537}{64}(a b c)^{3}=1-(a b)^{3}-c^{3}\left(a^{3}+b^{3}\right)-\frac{537}{64}(a b c)^{3}= \\
1-(a b)^{3}-c^{3}(a+b)^{3}+3 a b c^{3}(a+b)-\frac{537}{64}(a b c)^{3}= \\
1-(a b)^{3}-c^{3}(2-c)^{3}-\frac{... | not found | Inequalities | proof | Yes | Yes | inequalities | false | 734,293 |
Example 40 (Self-created, 2007.02.28) Let \( x, y, z \in \mathbf{R}^{-} \), then
$$27\left(\sum x^{3}\right)^{2} \geqslant\left(\sum x\right)^{4} \cdot \sum(-x+y+z)^{2}$$
Equality in (61) holds if and only if \( x=y=z \). | $$\begin{array}{l}
27\left(\sum x^{3}\right)^{2}-\left(\sum x\right)^{4} \cdot \sum(-x+y+z)^{2}= \\
3\left(3 \sum x^{3}-\sum x \cdot \sum x^{2}+\sum x \cdot \sum x^{2}\right)^{2}- \\
\left(\sum x\right)^{2}\left(\sum x^{2}+2 \sum y z\right) \cdot\left(3 \sum x^{2}-2 \sum y z\right)= \\
3\left(3 \sum x^{3}-\sum x \cdot ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,294 |
Example 41 Let $0 \leqslant a, b, c \leqslant 1$, then
$$\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}+(1-a)(1-b)(1-c) \geqslant \frac{7}{8}$$
Equality in (65) holds if and only if $a=b=c=\frac{1}{2}$. | $$\begin{array}{l}
\text { The left side of equation (65) - the right side }=\sum \frac{a}{b+c+1}+1-\sum a+\sum b c-a b c-\frac{7}{8}= \\
\left(\frac{1}{2} \sum b c-\frac{3}{4} \sum a+\sum \frac{a}{b+c+1}\right)+\left(\frac{1}{8}-\frac{1}{4} \sum a+\frac{1}{2} \sum b c-a b c\right)= \\
{\left[\frac{1}{4} \sum a(b+c+1)-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,295 |
Example 43 (Self-created, 2007.03.05) Let \(a, b, c, d \in \mathbf{R}^{-}\), and \(a^{4}+b^{4}+c^{4}+d^{4}=4\), then
$$(b c d)^{5}+(a c d)^{5}+(a b d)^{5}+(a b c)^{5} \leqslant 4$$
Equality in (67) holds if and only if \(a=b=c=d=1\). | Let $a^{4}=x, b^{4}=y, c^{4}=z, d^{4}=w$, and $x+y+z+w=4$, then
$$\begin{aligned}
4 \sum(b c d)^{5}= & 4 \sum\left(y z w \cdot y^{\frac{1}{4}} z^{\frac{1}{4}} w^{\frac{1}{4}}\right) \leqslant \sum y z w(y+z+w+1)= \\
& \sum x \cdot \sum y z w-4 x y z w+\sum y z w= \\
& \sum x \cdot \sum y z w-4 x y z w+\frac{1}{4} \sum ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,297 |
Example 19 (Self-created problem, 2007.01.21) Let $a, b, c, k \in \mathbf{R}$, then
$$\left(k^{2}+a^{2}\right)\left(k^{2}+b^{2}\right)\left(k^{2}+c^{2}\right) \geqslant 2 k^{3}\left(a^{2} b+b^{2} c+c^{2} a+a b c\right)$$
Equality in (31) holds if and only if $k=a=b=c$ or $k=0, a b c=0$. | Prove that the left side - right side $=\left(k^{3}-a b c\right)^{2}+k^{2} a^{2}(k-b)^{2}+k^{2} b^{2}(k-c)^{2}+$ $k^{2} c^{2}(k-a)^{2} \geqslant 0$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,298 |
Example 44 (Self-created, 07.03.03) Let $a, b, c \in \overline{\mathbf{R}^{-}}$, and $a+b+c=3$, then
$$\left(1+a^{3}\right)\left(1+b^{3}\right)\left(1+c^{3}\right) \geqslant 8$$
Equality in (70) holds if and only if $a=b=c=1$. | To prove that when one of $a, b, c$ is not less than 2, inequality (70) obviously holds. Therefore, we only need to discuss the case where $a, b, c$ are all less than 2, and inequality (70) holds.
Given $a+b+c=3$, we can assume $\frac{b+c}{2} \leqslant 1, 1 \leqslant a \leqslant 2$. We first prove
$$\left(1+b^{3}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,299 |
Example 46 (2007.05.10, Conjecture proposed by Chen Shengli, Fujian) Let $a_{1}, a_{2}, \cdots, a_{n} \in \mathbf{R}^{+}$, then
$$\sum_{i=1}^{n} a_{i} \cdot \sum_{i=1}^{n} \frac{1}{a_{i}} \geqslant n^{2}+\frac{4(n-1) \sum_{1<i<j<n}\left(a_{i}-a_{j}\right)^{2}}{\left(\sum_{i=1}^{n} a_{i}\right)^{2}}$$
Equality in (75) ... | By symmetry, without loss of generality, assume $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{k}$, and let $s=\sum_{i=1}^{n} a_{i}$. Then, equation (75) is equivalent to
$$\begin{array}{l}
s^{2} \sum_{1 \leqslant i<j<n} \frac{\left(a_{i}-a_{j}\right)^{2}}{a_{i} a_{j}}-4(n-1) \sum_{1<i<j<n}\left(a_{i}-a_{j}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,301 |
Example 47 (2007 Ukrainian Competition Problem) Let $a, b, c > 0$, and $abc \geqslant 1$, then
i )
$$\prod\left(a+\frac{1}{a+1}\right) \geqslant \frac{27}{8}$$
ii )
$$27 \prod\left(a^{3}+a^{2}+a+1\right) \geqslant 64 \prod\left(a^{2}+a+1\right)$$
Equality holds in both (77) and (78) if and only if $a=b=c=1$. | i) Since
$$2\left(a^{2}+a+1\right)^{2}-9 a\left(a^{2}+1\right)=(a-1)^{2}\left(2 a^{2}-a+2\right) \geqslant 0$$
There are also two similar inequalities, therefore
$$\Pi\left(a^{2}+a+1\right)^{2} \geqslant\left(\frac{9}{2}\right)^{3} a b c \Pi\left(a^{2}+1\right) \geqslant\left(\frac{9}{4}\right)^{3} \Pi(a+1)^{2}$$
Thu... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,302 |
Example 48 (Adapted Question, 2007.09.28) Let $a_{i}, b_{i} \in \mathbf{R}, i=1,2, \cdots, n, n$ be an odd number, and $\min \left\{a_{1}, a_{2}, \cdots, a_{n}\right\} \leqslant \min \left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$, and $\sum_{i=1}^{n} a_{i}^{k}=\sum_{i=1}^{n} b_{i}^{k}, k=1,2, \cdots, n-1$, then
$$\max \le... | Let $A_{1}, A_{2}, \cdots, A_{n}$ be the elementary symmetric polynomials of $a_{1}, a_{2}, \cdots, a_{n}$ (i.e., the sum of the products of $k(k=1,2, \cdots, n)$ distinct elements chosen from these $n$ numbers, the same applies below), and $B_{1}, B_{2}, \cdots, B_{n}$ be the elementary symmetric polynomials of $b_{1}... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,303 |
2. Let $\triangle A B C$ have sides $B C=a, C A=b, A B=c$, circumradius $R$, inradius $r$, and let $P$ be any point inside or on the boundary of $\triangle A B C$. The distances from point $P$ to the vertices $A, B, C$ are $R_{1}, R_{2}, R_{3}$, and the distances from $P$ to the sides $B C, C A, A B$ are $r_{1}, r_{2},... | 2. Note: This is an inequality proposed and proved by Mr. Chu Xiaoguang. For reference, see "Journal of Hunan Institute of Science and Technology", December 2003, Vol. 16, No. 4, by Chu Xiaoguang and Xiao Zhengang: "Proof of Several Geometric Inequality Conjectures".
Alternative Proof:
From \(a=\frac{a}{2 \Delta} \sum... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,304 |
3. (Recorded from Vasile Cirtoaje, editor, *Algebraic Inequalities. Old and New Methods*, § 3.4. Solutions, Problem 17) Let $a, b, c, d \in \mathbf{R}^{+}$, and $a b c d=1$, then
$$\sum(a-1)(a-2) \geqslant 0$$
with equality if and only if $a=b=c=d=1$. | 3. Proof: Let $s_{1}=a+b+c+d, s_{2}=ab+ac+ad+bc+bd+cd, s_{3}=bcd+acd+abd+abc, s_{4}=abcd=1$, then we have
$$s_{1}^{4}-4 s_{1}^{2} s_{2}+9 s_{1} s_{3}-16 s_{4} \geqslant 0$$
(see Example 14 in the first chapter "Proving Inequalities by Equivalent Transformation") and
$$s_{2}^{2}-3 s_{1} s_{3}+12 s_{4} \geqslant 0$$
(see... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,305 |
4. Let $a, b, c \in \mathbf{R}^{+}, \alpha \in \mathbf{R}$, then
$$a b c\left(a^{\alpha}+b^{\alpha}+c^{\alpha}\right) \geqslant a^{\alpha+2}(-a+b+c)+b^{\alpha+2}(a-b+c)+c^{\alpha+2}(a+b-c)$$
Equality holds if and only if $a=b=c$. | 4. Proof: Without loss of generality, assume $a \geqslant b \geqslant c$, then
left - right $=a^{\alpha+1}(a-b)^{2}+(a-b)(b-c)\left(a^{\alpha+1}-b^{\alpha+1}+c^{\alpha+1}\right)+$
$$c^{\alpha+1}(b-c)^{2} \geqslant(a-b)(b-c)\left(a^{\alpha+1}-b^{\alpha+1}+c^{\alpha+1}\right)$$
If $1+\alpha \geqslant 0$, then $a^{\alph... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,306 |
5. (Original problem, 1990. 12. 23) Let $x_{1}, x_{2}, x_{3}, x_{4} \in \mathbf{R}^{+}$, then
(1) $\frac{x_{1}}{x_{2}+x_{3}}+\frac{x_{2}}{x_{3}+x_{1}}+\frac{x_{3}}{x_{1}+x_{2}} \geqslant \frac{1}{2}\left(\frac{x_{1}+x_{2}}{x_{2}+x_{3}}+\frac{x_{2}+x_{3}}{x_{3}+x_{1}}+\frac{x_{3}+x_{1}}{x_{1}+x_{2}}\right)$;
(2) $\frac{... | 5. Proof:
(1) $2\left(\frac{x_{1}}{x_{2}+x_{3}}+\frac{x_{2}}{x_{3}+x_{1}}+\frac{x_{3}}{x_{1}+x_{2}}\right)-\left(\frac{x_{1}+x_{2}}{x_{2}+x_{3}}+\frac{x_{2}+x_{3}}{x_{3}+x_{1}}+\frac{x_{3}+x_{1}}{x_{1}+x_{2}}\right)=$ $\sum \frac{x_{1}-x_{2}}{x_{2}+x_{3}}=\sum\left(\frac{x_{1}-x_{2}}{x_{2}+x_{3}}+1\right)-3=\sum \frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,307 |
6. (Original problem, 1987.01.26) $P$ is a fixed point inside the tetrahedron $ABCD$. Any plane passing through $P$ divides the volume of the tetrahedron into two parts. The smallest possible volume of one part must be a tetrahedron. If $V_{PBCD}=v_{1}, V_{PCDA}=v_{2}, V_{PDAB}=v_{3}, V_{PABC}=v_{4}$, and $v_{1} \geqsl... | 7. Proof: $P, Q, R$ have two distributions.
When $P, Q, R$ are located on sides $B C, C A, A B$ respectively, and the areas of $\triangle Q A P, \triangle R B P, \triangle P C Q$ are denoted as $s_{1}, s_{2}, s_{3}, \frac{1}{3}(a+b+c)=k$. Also, let $B P=x$, then
$$\begin{array}{c}
s_{1}=\frac{(-x+a+b-k)(x+c-k)}{b c} s... | proof | Geometry | math-word-problem | Yes | Yes | inequalities | false | 734,308 |
Example 20 (Self-created problem, 2007.09.13) Let $a, b, c \in \overline{\mathbf{R}^{-}}$, and $a+b+c=2, n$ be a positive integer no less than 2, then
$$\sum \frac{1}{1+a^{n}} \geqslant 2$$
Equality in (32) holds if and only if one of $a, b, c$ is zero and the other two are equal to 1. | To prove that the original expression, after removing the denominator and rearranging, is
$$\sum b^{n} c^{n}+2(a b c)^{n} \leqslant 1$$
Given $a+b+c=2$, it is easy to see that $b c \leqslant 1, c a \leqslant 1, a b \leqslant 1$, thus we have
$$\sum b^{n} c^{n}+2(a b c)^{n} \leqslant \sum b^{n-1} c^{n-1}+2 a^{n-1} b^{n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,309 |
8. (Original problem, 2004.07.19) Let $x, y, z \in \mathbf{R}^{+}$, then
$$\frac{x^{2}}{y}+\frac{y^{2}}{z}+\frac{z^{2}}{x} \geqslant \sqrt[3]{9 \sum x^{3}}$$
Equality holds if and only if $x=y=z$. | 8. Simplified Proof
$$\begin{array}{l}
\left(\sum \frac{x^{2}}{y}\right)^{3}-9 \sum x^{3}=\left[\sum x+\sum \frac{(x-y)^{2}}{y}\right]^{3}-9 \sum x^{3} \geqslant \\
\left(\sum x\right)^{3}+3\left(\sum x\right)^{2} \cdot \sum \frac{(x-y)^{2}}{y}-9 \sum x^{3} \geqslant \\
\quad\left(\sum x\right)^{3}+3 \sum\left[y^{2}+2 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,310 |
9. (Original problem, 2000.11.11) Let \(a_{1}, a_{2}, \cdots, a_{n} \in \mathbf{R}^{+}, n \geqslant 3\), and define
$$\begin{array}{c}
A_{1}=1+a_{1}+a_{1} a_{2}+a_{1} a_{2} a_{3}+\cdots+a_{1} a_{2} \cdots a_{n-1} \\
A_{2}=1+a_{2}+a_{2} a_{3}+a_{2} a_{3} a_{4}+\cdots+a_{2} a_{3} \cdots a_{n} \\
A_{3}=1+a_{3}+a_{3} a_{4}... | 9. Proof: Note that $a_{i} A_{i+1}-A_{i}=a_{1} a_{2} \cdots a_{n}-1\left(i=1,2, \cdots, n, A_{n+1}=A_{1}\right)$, thus we have
$$\sum_{i=1}^{n} \frac{a_{i}}{A_{i}}=\sum_{i=1}^{n} \frac{1}{A_{1}}+\left(a_{1} a_{2} \cdots a_{n}-1\right)\left(\frac{1}{A_{1} A_{2}}+\frac{1}{A_{2} A_{3}}+\cdots+\frac{1}{A_{n} A_{1}}\right)$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,311 |
10. (Original Question, 2000. 11. 11) Let $p, q, r, x, y, z, x^{\prime}, y^{\prime}, z^{\prime} \in \overline{\mathbf{R}^{-}}, \lambda, u, v \in \mathbf{R}^{+}$, and
$$\left\{\begin{array}{l}
2 p \geqslant \frac{u}{\lambda} z+\frac{v}{\lambda} y \\
2 q \geqslant \frac{v}{u} x+\frac{\lambda}{u} z^{\prime} \\
2 r \geqsla... | 10. It is easy to prove (1) and (3).
(2)
$$\begin{aligned}
4 \sum q r= & \left(\frac{u}{v} y^{\prime} z+\frac{v}{u} y z^{\prime}\right)+\left(\frac{\lambda}{v} x^{\prime} z^{\prime}+\frac{v}{\lambda} x z\right)+\left(\frac{\lambda}{u} x y^{\prime}+\frac{u}{\lambda} x^{\prime} y\right)+ \\
& x x^{\prime}+y y^{\prime}+z ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,312 |
11. (2002 IMO Shortlist) Let $x_{i} \in \mathbf{R}(i=1,2, \cdots, n)$, prove that
$$\frac{x_{1}}{1+x_{1}^{2}}+\frac{x_{2}}{1+x_{1}^{2}+x_{2}^{2}}+\cdots+\frac{x_{n}}{1+x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}<\sqrt{n}$$ | 11. Proof
$$\text { Original } \Leftrightarrow\left(\frac{x_{1}}{1+x_{1}^{2}}+\frac{x_{2}}{1+x_{1}^{2}+x_{2}^{2}}+\cdots+\frac{x_{n}}{1+x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}\right)^{2}<n$$
By the Cauchy-Schwarz inequality, we have
$$\text { Left side of the above } \leqslant n\left[\frac{x_{1}^{2}}{\left(1+x_{1}^{2}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,313 |
12. (Adapted Question, 2005.01.01) If $x, y, z \in \mathbf{R}$, and $x^{2}+y^{2}+z^{2} \leqslant 2$, prove that
(1) When at least one of $x, y, z$ is not less than zero, we have $2 x y z + 1 \geqslant \sum y z$;
(2) $\sum y z + 3 \geqslant 2 \sum x$.
Equality holds in both inequalities if and only if one of $x, y, z$ ... | 12. Proof: (1) First, prove
$$2+x y z \geqslant \sum x$$
Since
$$1-y z \geqslant \frac{1}{2} \sum x^{2}-y z=\frac{1}{2} x^{2}+\frac{1}{2}(y-z)^{2} \geqslant 0$$
Similarly, we have
$$1-z x \geqslant 0,1-x y \geqslant 0$$
Therefore,
$$\begin{aligned}
2^{2}-\left(\sum x-x y z\right)^{2}= & 4-\sum x^{2}-2 \sum y z+2 x y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,314 |
13. (Original problem, 2007.02.03) Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5} \in \mathbf{R}^{-}$, prove that
$$\prod_{i=1}^{5} \frac{1+a_{i}^{3}}{1+a_{i}^{2}} \geqslant \frac{1+a_{1} a_{2} a_{3} a_{4} a_{5}}{2}$$
Equality holds if and only if $a_{1}=a_{2}=a_{3}=a_{4}=a_{5}$. | 13. Proof: Since
$$\begin{array}{l}
2\left(1+a_{i}^{3}\right)^{5}-\left(1+a_{i}^{2}\right)^{5}\left(1+a_{i}^{5}\right)= \\
(1-a)^{4}\left(1+4 a+5 a^{2}+10 a^{3}+15 a^{5}+15 a^{6}+15 a^{7}+10 a^{8}+\right. \\
\left.5 a^{9}+4 a^{10}+a^{11}\right) \geqslant 0 \\
\quad 2\left(1+a_{i}^{3}\right)^{5} \geqslant\left(1+a_{i}^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,315 |
14. (Original problem, 2001.09.09) Let $x, y, z, a, b, c \in \mathbf{R}^{+}$, and $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=\lambda \leqslant 1$, prove that
$$\frac{x}{a}+\frac{y}{b}+\frac{z}{c} \leqslant 1+\frac{9 \lambda^{2}-\lambda^{3}}{27} \cdot \frac{x y z}{a b c}$$ | 14. Proof
So
$$\begin{aligned}
3-\lambda= & \frac{x-a}{x}+\frac{y-b}{y}+\frac{z-c}{z} \geqslant \\
& 3 \sqrt[3]{\frac{(x-a)(y-b)(z-c)}{x y z}}
\end{aligned}$$
$$(x-a)(y-b)(z-c) \leqslant\left(\frac{3-\lambda}{3}\right)^{3} x y z$$
It is also easy to prove
$$\begin{aligned}
\frac{x}{a}+\frac{y}{b}+\frac{z}{c}= & 1+\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,316 |
15. Let $a, b, c \in \mathbf{R}^{-}, n \geqslant 2$, and $n$ is a positive integer, then
$$\sum \sqrt[n]{\frac{a}{b+c}} \geqslant 2$$
Equality holds if and only if one of $a, b, c$ is zero and the other two are equal. | 15. Proof: It suffices to prove a local inequality:
$$\sqrt[n]{\frac{a}{b+c}} \geqslant \frac{2 a^{\frac{2}{n}}}{a^{\frac{2}{n}}+b^{\frac{2}{n}}+c^{\frac{2}{n}}}$$
Let \( x = a^{\frac{1}{n}}, y = b^{\frac{1}{n}}, z = c^{\frac{1}{n}} \), then we need to prove
Also,
$$\begin{array}{c}
\sqrt[n]{\frac{x^{n}}{y^{n}+z^{n}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,317 |
16. Let $a, b, c > 1$, and $\sum \frac{1}{a^{2}-1}=1$, then
$$\sum \frac{1}{a+1} \leqslant 1$$
Equality holds if and only if $a=b=c=2$. | 16. Proof
$$\begin{aligned}
\frac{1}{1+a}-\frac{1}{3} \leqslant & \frac{1}{4}\left(\frac{1}{a^{2}-1}-\frac{1}{3}\right) \Leftrightarrow \\
& (2-a)\left(a^{2}-1\right) \leqslant \\
& \frac{1}{4}(2-a)(2+a)(a+1) \Leftrightarrow \\
& \frac{3}{4}(a+1)(2-a)^{2} \geqslant 0
\end{aligned}$$
Thus, the original inequality is pr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,318 |
17. Let $a, b, c \in \mathbf{R}^{+}$, then
$$9 a(a+b)(a+b+c) \geqslant 2(a+\sqrt{a b}+\sqrt[3]{a b c})^{3}$$
Equality holds if and only if $a=b=c$. | 17. Proof: Since $\frac{a+b}{2} \geqslant \frac{a}{3}+\frac{\sqrt{a b}}{3}+\frac{b}{3}$
Therefore
$$\begin{aligned}
a(a+b)(a+b+c) \geqslant & 6\left(\frac{a}{3}+\frac{a}{3}+\frac{a}{3}\right)\left(\frac{a}{3}+\frac{\sqrt{a b}}{3}+\frac{b}{3}\right)\left(\frac{a}{3}+\frac{b}{3}+\frac{c}{3}\right) \geqslant \\
& 6\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,319 |
Example 21 (from "Math Sakura Station Forum — Southern Subject Network Forum" in the "Inequality" section, December 23, 2006, proposed by hangjingjun) Let $x, y, z \in \overline{\mathbf{R}^{-}}$, and $\sum y z=1$, then
$$\sum\left(x^{2}+\frac{4}{1+x^{2}}\right) \geqslant 10$$
Equality holds in (33) if and only if $x=y... | To prove that the homogeneous form of equation (33) is equivalent to the following:
$$\sum x^{2}+\frac{8 \sum x \cdot\left(\sum y z\right)^{2}}{\prod(y+z)}-10 \sum y z \geqslant 0$$
Let \( s_{1}=\sum x, s_{2}=\sum y z, s_{3}=x y z \), then the above equation, after removing the denominator and simplifying, can be writ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,320 |
18. Let $x_{i}, y_{i} \in \mathbf{R}(i=1,2, \cdots, n)$, and $\sum_{i=1}^{n} x_{i} \geqslant 0, \sum_{i=1}^{n} y_{i} \geqslant 0, \sum_{1<i<j<n} x_{i} x_{j} \geqslant 0$, $\sum_{1<i<j<n} y_{i} y_{j} \geqslant 0$, denote $\sum y_{1}=y$, then
the equality holds if and only if $\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}=\cd... | 18. Simplified Proof
But
$$\begin{array}{l}
\text { Original } \Leftrightarrow \sum x_{1} \cdot \sum y_{1}\left(\sum x_{1} \sum y_{1}-\sum x_{1} y_{1}\right) \geqslant \\
\left(\sum x_{1}\right)^{2} \cdot \sum_{1<i<j<n} y_{i} y_{j}+\left(\sum y_{1}\right)^{2} \sum_{1<i<j<n} x_{i} x_{j} \Leftrightarrow \\
\left(\sum x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,321 |
21. In $\triangle A B C$, the altitudes to sides $B C, C A, A B$ are $h_{a}, h_{b}, h_{c}$, respectively, and $B C=a, C A=b$, $A B=c$. Prove that
$$\frac{9}{4} a^{2} \geqslant 2 h_{b}^{2}+2 h_{c}^{2}-h_{a}^{2}$$ | 21. Simplified Proof
$$\begin{aligned}
\text { Original } \Leftrightarrow & 9 a^{2}+3 \cdot \frac{16 \Delta^{2}}{a^{2}} \geqslant 32 \Delta^{2} \cdot \sum \frac{1}{a^{2}} \Leftrightarrow \\
& 3\left[3 a^{2}+\frac{2 a^{2}\left(b^{2}+c^{2}\right)-\left(b^{2}-c^{2}\right)^{2}-a^{4}}{a^{2}}\right] \geqslant \\
& 2\left(2 \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,322 |
22. In $\triangle A B C$, $B C=a, C A=b, A B=c$, and the medians to these sides are $m_{a}, m_{b}$, $m_{c}$, respectively, then
$$a b c m_{a} m_{b} m_{c} \geqslant \frac{\Delta}{2} \sum b^{2} c^{2}$$
Equality holds if and only if $\triangle A B C$ is an isosceles triangle. | 22. Proof
$$\begin{array}{l}
\prod\left(2 a m_{a}\right)^{2}=\prod\left[16 \Delta^{2}+\left(b^{2}-c^{2}\right)\right]=\left(16 \Delta^{2}\right)^{3}+\left(16 \Delta^{2}\right)^{2} \cdot \sum\left(b^{2}-c^{2}\right)^{2}+ \\
16 \Delta^{2} \sum\left(a^{2}-b^{2}\right)^{2}\left(b^{2}-c^{2}\right)^{2}+\left(b^{2}-c^{2}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,323 |
23. (Original problem, 1998.04.10) In $\triangle A B C$, $B C=a, C A=b, A B=c$, and the medians to the corresponding sides are $m_{a}, m_{b}, m_{c}$, respectively. Prove that
(1) $\sum a m_{a} \geqslant \sqrt{\sum b^{2} c^{2}}+2 \Delta$;
(2) $\sum \frac{1}{a m_{a}} \geqslant \frac{1}{\sqrt{\sum b^{2} c^{2}}}+\frac{1}{2... | 23. Proof
$$\begin{array}{l}
\text { Equation }(1) \Leftrightarrow \sum\left(2 a m_{a}-4 \Delta\right) \geqslant 2\left(\sqrt{\sum b^{2} c^{2}}-4 \Delta\right) \Leftrightarrow \\
\sum \frac{\left(2 a m_{a}\right)^{2}-16 \Delta^{2}}{2 a m_{a}+4 \Delta} \geqslant \frac{2\left[\sum b^{2} c^{2}-16 \Delta^{2}\right]}{\sqrt{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,324 |
24. (Original Question, 2000.10.13) Let $a, b, c$ be the lengths of the sides of $\triangle ABC$, then
$$a^{2} b+b^{2} c+c^{2} a \leqslant \frac{1}{8}\left(\sum a\right)^{3}-\frac{3}{8} \prod(-a+b+c)$$ | 24. Simplify
$$\begin{array}{l}
\left(\sum a\right)^{3}-8\left(a^{2} b+b^{2} c+c^{2} a\right)=\left[\sum a^{3}-\sum a b(a+b)+2 a b c\right]+ \\
4\left(a b^{2}+b c^{2}+c a^{2}-a^{2} b-b^{2} c-c^{2} a\right)+4 a b c= \\
-\prod(-a+b+c)-4(a-b)(b-c)(a-c)+4 a b c \geqslant \\
-\prod(-a+b+c)-4(a-b)(b-c)(a-c)+ \\
4 \prod(-a+b+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,325 |
25. In $\triangle A B C$, the lengths of the three sides are $B C=a, C A=b, A B=c$, the median to side $B C$ is $m_{a}$, and the angle bisectors of $\angle B$ and $\angle C$ are $t_{b}$ and $t_{c}$, respectively. Then,
$$m_{a}+t_{b}+t_{c} \leqslant \frac{\sqrt{3}}{2}(a+b+c)$$ | 25. Proof
$$\begin{aligned}
m_{a}+2 t_{b} \leqslant & \sqrt{3\left(m_{a}^{2}+2 t_{b}^{2}\right)} \leqslant \sqrt{\frac{3}{4}\left[2 b^{2}+2 c^{2}-a^{2}+2(a+c)^{2}-2 b^{2}\right]}= \\
& \frac{\sqrt{3}}{2}(a+2 c)
\end{aligned}$$
Similarly, \( m_{a}+2 t_{c} \leqslant \frac{\sqrt{3}}{2}(a+2 b) \). Adding the two inequalit... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,326 |
26. (Original problem, 1989.07.16) Let $\alpha, \beta, \gamma, \theta \in \mathbf{R}$, and $\alpha+\beta+\gamma+\theta=\pi$. In a plane convex quadrilateral $A B C D$, we have
$$\frac{\sin \alpha}{\sin \frac{A}{2}}+\frac{\sin \beta}{\sin \frac{B}{2}}+\frac{\sin \gamma}{\sin \frac{C}{2}}+\frac{\sin \theta}{\sin \frac{D}... | 26. Prompt: In Chapter 1, "Proving Inequalities by Equivalent Transformation Method," in formula (9), let \( x = \frac{1}{\sin \frac{A}{2}} \), \( y = \frac{1}{\sin \frac{B}{2}} \), \( z = \frac{1}{\sin \frac{C}{2}} \), \( w = \frac{1}{\sin \frac{D}{2}} \). At this point, it can be proven that
$$
\sqrt{\frac{(x y + z w... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,327 |
27. (Original problem, 1993.05.05) In a planar convex quadrilateral $A B C D$, we have
$$\begin{array}{l}
\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)\left(\tan \frac{B}{2}+\tan \frac{C}{2}\right)\left(\tan \frac{C}{2}+\tan \frac{D}{2}\right)\left(\tan \frac{D}{2}+\tan \frac{A}{2}\right) \geqslant \\
\left(\tan \frac... | 27. (Original problem, 1993.05.05) In a plane convex quadrilateral $ABCD$, we have
$$\begin{array}{l}
\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)\left(\tan \frac{B}{2}+\tan \frac{C}{2}\right)\left(\tan \frac{C}{2}+\tan \frac{D}{2}\right)\left(\tan \frac{D}{2}+\tan \frac{A}{2}\right) \geqslant \\
\left(\tan \frac{A}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,328 |
28. $P, Q$ are any two points on the plane of $\triangle A_{1} A_{2} A_{3}$, then
$$\begin{array}{ll}
& P A_{1} \cdot Q A_{1} \sin A_{1}+P A_{2} \cdot Q A_{2} \sin A_{2}+P A_{3} \cdot Q A_{3} \sin A_{3} \geqslant 2 \Delta \\
\text { or } & P A_{1} \cdot Q A_{1} \cdot a_{1}+P A_{2} \cdot Q A_{2} \cdot a_{2}+P A_{3} \cd... | 28. $P, Q$ are any two points on the plane of $\triangle A_{1} A_{2} A_{3}$, then
or
$$\begin{array}{l}
P A_{1} \cdot Q A_{1} \sin A_{1}+P A_{2} \cdot Q A_{2} \sin A_{2}+P A_{3} \cdot Q A_{3} \sin A_{3} \geqslant 2 \Delta \\
P A_{1} \cdot Q A_{1} \cdot a_{1}+P A_{2} \cdot Q A_{2} \cdot a_{2}+P A_{3} \cdot Q A_{3} \cdo... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,329 |
Example 22 Let $a, b, c \in \mathbf{R}^{+}$, and $x=a+\frac{1}{b}-1, y=b+\frac{1}{c}-1, z=c+\frac{1}{a}-1$, then
$$x y+y z+z x \geqslant 3$$
The equality in (34) holds if and only if $x=y=z=1$. | To prove that
$$\begin{aligned}
x y+y z+z x-3= & \left(a+\frac{1}{b}-1\right)\left(b+\frac{1}{c}-1\right)+\left(b+\frac{1}{c}-1\right)\left(c+\frac{1}{a}-1\right)+ \\
& \left(c+\frac{1}{a}-1\right)\left(a+\frac{1}{b}-1\right)-3= \\
& \frac{1}{a b c}[a(a b-b+1)(b c-c+1)+ \\
& b(b c-c+1)(c a-a+1)+ \\
& c(c a-a+1)(a b-b+1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,332 |
32. (Original problem, 1985.03.16) Let $A B C D$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ be two convex planar quadrilaterals, and denote $A B=a, B C=b, C D=c, D A=d, A^{\prime} B^{\prime}=a^{\prime}, B^{\prime} C^{\prime}=b^{\prime}, C^{\prime} D^{\prime}=c^{\prime}, D^{\prime} A^{\prime}=d^{\prime}$, with ar... | 32. Prompt: There is an identity
$$\begin{aligned}
K^{2}= & \left(16 F F^{\prime}\right)^{2}+\left[2(a b+c d)\left(a^{\prime 2}+b^{\prime 2}-c^{\prime 2}-d^{\prime 2}\right)-\right. \\
& \left.2\left(a^{\prime} b^{\prime}+c^{\prime} d^{\prime}\right)\left(a^{2}+b^{2}-c^{2}-d^{2}\right)\right]^{2}
\end{aligned}$$
or ac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,335 |
34. (Original problem, 1991. 10.02) Let $y_{1}, y_{2}, y_{3}, y_{4} \in \mathbf{R}$, and $\sum_{i=1}^{4} \frac{1}{1+y_{i}}=2, \prod_{i=1}^{4}(1+$ $\left.y_{i}\right)>0$, prove that
$$2\left(y_{1} y_{2}+y_{1} y_{3}+y_{1} y_{4}+y_{2} y_{3}+y_{2} y_{4}+y_{3} y_{4}\right) \geqslant 3\left(y_{1}+y_{2}+y_{3}+y_{4}\right)$$
... | 34. Proof: Let $x_{i}=\frac{1}{1+y_{i}}$, i.e., $y_{i}=\frac{1-x_{i}}{x_{i}}=\frac{\sum x_{1}-2 x_{i}}{2 x_{i}}$, then
$$\begin{array}{l}
\text { Original } \Leftrightarrow 2\left[\frac{\left(x-2 x_{1}\right)\left(x-2 x_{2}\right)}{4 x_{1} x_{2}}+\frac{\left(x-2 x_{1}\right)\left(x-2 x_{3}\right)}{4 x_{1} x_{3}}+\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,337 |
35. (Original problem, 1992. 04. 11) Let $a_{i} \in \mathbf{R}$, and $| a_{i} | \leqslant 1 (i=1,2, \cdots, n)$, and let $s_{k}$ $(k \in \mathbf{N}, k \leqslant n)$ denote the sum of the products of every $k$ numbers among $a_{1}, a_{2}, \cdots, a_{n}$, then
$$s_{2}+s_{4}+s_{6}+\cdots+\left\{\begin{array}{l}
s_{n}(n \t... | 35. Hint: $\prod_{i=1}^{n}\left(1+a_{i}\right)+\prod_{i=1}^{n}\left(1-a_{i}\right) \geqslant 0$. Expand to obtain. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,338 |
36. (Original problem, 1993. 10.01) In $\triangle A B C$, the circumradius is $R$, the inradius is $r$, and the angle bisectors of $\angle A, \angle B, \angle C$ are $w_{a}, w_{b}, w_{c}$, respectively, then
$$\sum \frac{1}{w_{0}} \geqslant \frac{1}{R}+\frac{1}{2 r}$$
Equality holds if and only if $\triangle A B C$ is... | 36. Proof: It involves using $4 a^{2} b c \geqslant(b+c)^{2}(a-b+c)(a+b-c)$ (which is easy to prove). Thus,
$$\begin{aligned}
w_{b} w_{c}= & \frac{\sqrt{a c(a+b+c)(a-b+c)}}{a+c} \cdot \frac{\sqrt{a b(a+b+c)(a+b-c)}}{a+b}= \\
& \frac{(a+b+c) \cdot \sqrt{4 a^{2} b c} \cdot \sqrt{(a-b+c)(a+b-c)}}{2(a+b)(a+c)} \geqslant \\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,339 |
38. Let $F$ be the Fermat point of $\triangle ABC$, and the distances from point $F$ to sides $BC$, $CA$, and $AB$ are $r_{1}$, $r_{2}$, and $r_{3}$, respectively. The circumradius of $\triangle ABC$ is $R$, the inradius is $r$, and the area is $\Delta$. Then
(1) $r_{1}+r_{2}+r_{3} \leqslant 3 r$
(2) (Original problem,... | 38. Note: (1) The inequality in (1) was given by Mr. Wu Yuesheng from Zhejiang. See "Inequalities about Fermat Point" in *Mathematics in Middle School*, Issue 4, 1997. (2) The inequality in (2) was given by the author. The solution is found in *Mathematics in Middle School*, Issue 2, 1998, in the article "Another Inequ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,341 |
39. (Original Question, 1900.07.25) Let $a, b, c, d \in \mathbf{R}^{-}, a^{2}+b^{2}-k a b=m^{2}, c^{2}+d^{2}-$ $k c d=n^{2}$, then
$$(m-n)^{2} \geqslant \frac{(a b-c d)(a c-b d)(a d-b c)}{a b c d}$$
Equality holds if and only if $\frac{m}{a b}=\frac{n}{c d}$. | 39. Simplify
$$\frac{a^{2}+b^{2}-m^{2}}{a b}=\frac{c^{2}+d^{2}-n^{2}}{c d} \Leftrightarrow \frac{(a c-b d)(a d-b c)}{a b c d}=\frac{m^{2}}{a b}-\frac{n^{2}}{c d}$$
Therefore,
$$\frac{(a b-c d)(a c-b d)(a d-b c)}{a b c d}=(a b-c d)\left(\frac{m^{2}}{a b}-\frac{n^{2}}{c d}\right) \leqslant(m-n)^{2}$$
Note: The conditio... | (m-n)^{2} \geqslant \frac{(a b-c d)(a c-b d)(a d-b c)}{a b c d} | Inequalities | proof | Yes | Yes | inequalities | false | 734,342 |
For example, let $23 a, b, c$ and $x, y, z$ be non-negative real numbers, and $x+y+z=a+b+c$, then
$$a x^{2}+b y^{2}+c z^{2}+x y z \geqslant 4 a b c$$
Equality in (37) holds if and only if $x=-a+b+c, y=a-b+c, z=a+b-c$. | Given $a+b+c=x+y+z$, we know that among $b+c \geqslant x, a+c \geqslant y, a+b \geqslant z$, at least one inequality holds. Without loss of generality, assume $a+c \geqslant y$. If $y=a+c$, then $b=x+z$, thus
$$\begin{array}{l}
a x^{2}+b y^{2}+c z^{2}+x y z-4 a b c= \\
a x^{2}+(a+c)^{2}(x+z)+c z^{2}+ \\
(a+c) x z-4 a c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,343 |
40. (Original Question, 1990. 10.09) Let $a_{1}, a_{2}, b_{1}, b_{2} \in \mathbf{R}$, and real numbers $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ satisfy
$$\begin{array}{c}
\left(a_{1}^{2}+a_{2}^{2}\right) x^{2}+\left(b_{1}^{2}+b_{2}^{2}\right) y^{2}+2\left(a_{1} b_{1}+a_{2} b_{2}\right) x y=1 \\
\left|x_{1}... | 40. Hint: The original condition can be rewritten as
$$\begin{array}{c}
\left(a_{1} x+b_{1} y\right)^{2}+\left(a_{2} x+b_{2} y\right)^{2}=1 \\
a_{1} x+b_{1} y=\cos \alpha, a_{2} x+b_{2} y=\sin \alpha
\end{array}$$ | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,344 |
41. ("Math Olympiad Home Website", March 31, 2007, proposed by slsh, modified by the author) Let $P$, $A_{1}$, $A_{2}$, $A_{3}$ be any four points on a plane, then
$$A_{2} A_{3} \cdot P A_{1}^{2} + A_{3} A_{1} \cdot P A_{2}^{2} + A_{1} A_{2} \cdot P A_{3}^{2} \geqslant A_{2} A_{3} \cdot A_{3} A_{1} \cdot A_{1} A_{2}$$
... | 41. Proof: Using complex number methods to prove.
Let the complex numbers corresponding to $P, A_{1}, A_{2}, A_{3}$ be $z, z_{1}, z_{2}, z_{3}$, respectively, then
$$\begin{array}{l}
\left|A_{2} A_{3}\right| \cdot P A_{1}^{2}+\left|A_{3} A_{1}\right| \cdot P A_{2}^{2}+\left|A_{1} A_{2}\right| \cdot P A_{3}^{2}= \\
\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,345 |
42. (Original problem, 1988. 11.28) Let $\alpha_{i}, \beta_{i} \in(0, \pi), i=1,2,3,4, \alpha_{i}, \beta_{i}(i=1,2$, $3,4)$, where at most one of them is zero, and $\sum_{i=1}^{4} \alpha_{i}=\sum_{i=1}^{4} \beta_{i}=\pi$, then
$$\sum_{i=1}^{4} \sin \alpha_{i} \cos \beta_{i} \geqslant \frac{1}{2}\left[\left(\sum_{i=1}^{... | 42. Prompt
$$\begin{array}{l}
4 \sum \sin \alpha_{1} \cos \alpha_{1}= \\
\sum\left(-\sin \alpha_{1}+\sin \alpha_{2}+\sin \alpha_{3}+\sin \alpha_{4}\right)\left(-\cos \alpha_{1}+\cos \alpha_{2}+\cos \alpha_{3}+\cos \alpha_{4}\right) \geqslant \\
4\left[\pi\left(-\sin \alpha_{1}+\sin \alpha_{2}+\sin \alpha_{3}+\sin \alph... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,346 |
43. (Original problem, 1992.07.23) Let $x, y, a, b \in \mathbf{R}$, and $x^{2}+y^{2} \leqslant 1$, prove that
$$\left|a x^{2}+b x y+c y^{2}\right| \leqslant\left|\frac{a+c}{2}\right|+\sqrt{\left(\frac{a-c}{2}\right)^{2}+\left(\frac{b}{2}\right)^{2}}$$ | 43. Proof: Let $x=k \cos \alpha, y=k \sin \alpha(0 \leqslant k \leqslant 1)$, then
$$\begin{aligned}
\left|a x^{2}+b x y+c y^{2}\right|= & k^{2}\left|a \cos ^{2} \alpha+b \sin \alpha \cos \alpha+c \sin ^{2} \alpha\right| \leqslant \\
& k^{2}\left|\frac{a+c}{2}+\frac{a-c}{2} \cos 2 \alpha+\frac{b}{2} \sin 2 \alpha\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,347 |
44. (Original problem, 2006.06.08) Let $x, y, z \in \mathbf{R}$, then
$$6 \sum(x-y+z) x^{3} \geqslant 3 \sum y^{2} z^{2}+\left(\sum y z\right)^{2}$$ | 44. Simplify
\[
\begin{aligned}
6 \sum(x-y+z) x^{3}= & 3 \sum y^{2} z^{2}+3 \sum\left(-x^{2}+y^{2}+x y\right)^{2} \geqslant \\
& 3 \sum y^{2} z^{2}+\left(\sum y z\right)^{2}
\end{aligned}
\] | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,348 |
45. (Original problem, 2008.01.11) Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a \geqslant b>0)$, the line $B D$ passes through the left focus, intersecting the ellipse at points $B, D$, and the line $A C$ passes through the right focus, intersecting the ellipse at points $A, C$. The angle between line... | 45. Proof: (1) If the slopes of lines $B D, A C$ exist and are $k_{1}, k_{2}$ respectively, and the coordinates of $F_{1}, F_{2}$ are $(-c, 0), (c, 0)$, then the equations of lines $B D, A C$ are $y=k_{1}(x+c), y=k_{2}(x-c)$, and
$$\left|1+k_{1} k_{2}\right|=\left|k_{2}-k_{1}\right| \cot \alpha$$
Substituting $y=k_{1}... | proof | Geometry | proof | Yes | Yes | inequalities | false | 734,349 |
46. (Recorded from Vasile Cirtoaje, editor, *Algebraic Inequalities. Old and New Methods*, § 3.4. Applications, Problem 30) Let \( x, y, z \in \mathbf{R}^{+} \), and \( x+y+z=3 \), then
$$8\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+9 \geqslant 10\left(x^{2}+y^{2}+z^{2}\right)$$
Equality holds if and only if one ... | 46. Proof: Let
$$s_{1}=x+y+z=3, s_{2}=yz+zx+xy, s_{3}=xyz$$
Then
$$\text { Original expression } \Leftrightarrow 8 s_{2}-81 s_{3}+20 s_{2} s_{3} \geqslant 0$$
Let $w=\sqrt{s_{1}^{2}-3 s_{2}}=\sqrt{9-3 s_{2}}$, i.e., $s_{2}=\frac{9-w^{2}}{3}(0 \leqslant w \leqslant 3)$. By Theorem 2 in the appendix of Chapter 4 "Appli... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,350 |
48. Let $x, y, z \in \mathbf{R}^{+}$, prove that
$$\left(x^{2}+\frac{3}{4}\right)\left(y^{2}+\frac{3}{4}\right)\left(z^{2}+\frac{3}{4}\right) \geqslant \sqrt{\prod(y+z)}$$
Equality holds if and only if $x=y=z=\frac{1}{2}$. | 48. Hint: From
$$\begin{aligned}
\left(y^{2}+\frac{3}{4}\right)\left(z^{2}+\frac{3}{4}\right)-(y+z)= & \frac{1}{4}(y+z-1)^{2}+\left(y z-\frac{1}{4}\right)^{2}+ \\
& \frac{1}{2}\left(y-\frac{1}{2}\right)^{2}+\frac{1}{2}\left(z-\frac{1}{2}\right)^{2} \geqslant 0
\end{aligned}$$
we have
$$\left(y^{2}+\frac{3}{4}\right)\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,352 |
49. $a, b, c \in \mathbf{R}^{+}$, and $a b c=1$, prove that
$$\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)} \geqslant \frac{3}{4}$$
Equality holds if and only if $a=b=c=1$. | 49. Simplify
$$\begin{array}{l}
\sum \frac{a}{(a+1)(b+1)}-\frac{3}{4}=\frac{\sum bc+\sum a}{\prod(a+1)}-\frac{3}{4}= \\
\frac{4 \sum bc+4 \sum a-3-3 \sum a-3 \sum bc-3abc}{4 \prod(1+a)}= \\
\frac{\sum a+\sum bc-6}{4 \prod(a+1)} \geqslant 0
\end{array}$$
(Note that \(abc=1\), and \(\sum a \geqslant 3 \sqrt[3]{abc}=3\), ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,353 |
Example 24 (Self-created problem, 2006. 12.17) Let $a, b, c, d \in \mathbf{R}^{-}$, and $a c=1$, then
$$\frac{a}{a+a b+1}+\frac{b}{b+b c+1}+\frac{c}{c+c d+1}+\frac{d}{d+d a+1} \leqslant \frac{4}{3}$$
Equality in (38) holds if and only if $a=b=c=d=1$. | $$\begin{array}{l}
\frac{a}{a+a b+1}+\frac{b}{b+b c+1}+\frac{c}{c+c a+1}= \\
1-\frac{(1-a b c)^{2}}{(a+a b+1)(b+b c+1)(c+c a+1)}
\end{array}$$
This is because
$$\begin{aligned}
\text { The left side of the above equation }= & \frac{a}{a+a b+1}+\frac{a b}{a+a b+1}+\frac{1}{a+a b+1}+\frac{b}{b+b c+1}- \\
& \frac{a b}{a+... | \frac{4}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 734,354 |
50. $a, b, c$ are positive numbers, prove that
$$\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a} \geqslant a+b+c+\frac{4(a-b)^{2}}{a+b+c}$$
Equality holds if and only if $a>c>b$, and $\frac{a}{a-c}=\frac{b}{a-b}=\frac{c}{c-b}$, i.e., $a>c>b$ and $b^{3}+c^{3}=2 b^{2} c$. | 50. Proof: $\frac{a^{2}}{b}+b-2 a=\frac{(a-b)^{2}}{b}, \frac{b^{2}}{c}+c-2 b=\frac{(b-c)^{2}}{c}, \frac{c^{2}}{a}+a-2 c=$ $\frac{(c-a)^{2}}{a}$, adding the above three equations, we get
$$\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}=(a+b+c)+\frac{(a-c)^{2}}{a}+\frac{(a-b)^{2}}{b}+\frac{(c-b)^{2}}{c}$$
Applying the ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,355 |
51. (Original problem, 1988. 12.13) Let $x, y, z \in \overline{\mathbf{R}^{-}}$, and $x^{2}+y^{2}+z^{2} \leqslant 3$, then
$$27+36 x y z \geqslant 7 \sum x \sum y z$$
Equality holds if and only if $x=y=z=1$. | 51. Proof: Since
$$\begin{aligned}
27-12 \sum x+\sum x \sum y z= & 27-12 \sum x+\sum x \cdot \frac{\left(\sum x\right)^{2}-\sum x^{2}}{2} \geqslant \\
& 27-12 \sum x+\frac{\left(\sum x\right)^{3}-3 \sum x}{2}= \\
& \frac{1}{2}\left(\sum x-3\right)^{2}\left(\sum x+6\right) \geqslant 0
\end{aligned}$$
Therefore, it suff... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,356 |
52. Let $a, b, c>0$, and $a+b+c=1$, prove that
$$\sqrt{\frac{a b}{c+a b}}+\sqrt{\frac{b c}{a+b c}}+\sqrt{\frac{c a}{b+c a}} \leqslant \frac{3}{2}$$
Equality holds if and only if $a=b=c=1$. | 52. Simplify
$$\sqrt{\frac{a b}{c+a b}}=\sqrt{\frac{a b}{1-a-b+a b}}=\sqrt{\frac{a b}{(1-a)(1-b)}} \leqslant \frac{1}{2}\left(\frac{a}{1-b}+\frac{b}{1-a}\right)$$
Similarly, there are two other expressions, so
$$\begin{aligned}
\sqrt{\frac{a b}{c+a b}}+\sqrt{\frac{b c}{a+b c}}+\sqrt{\frac{c a}{b+c a}}= & \sum \sqrt{\f... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 734,357 |
53. Positive numbers $x, y, z$ satisfy $x+y+z+x^{2}+y^{2}+z^{2}=6$, prove that $x^{2} y+y^{2} z+z^{2} x \leqslant 3$ | 53. Proof: Since
$$6=\sum x+\sum x^{2} \geqslant 2 \sqrt{\sum x \sum x^{2}}$$
Therefore,
$$\begin{aligned}
9 \geqslant & \sum x \sum x^{2}=\sum x^{3}+\sum x^{2}(y+z)= \\
& \sum\left(x^{3}+x y^{2}\right)+\sum x^{2} y \geqslant \\
& 2 \sum x^{2} y+\sum x^{2} y=3 \sum x^{2} y
\end{aligned}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,358 |
54. (Original problem, 2007.07.08) Let $a, b, c \in \mathbf{R}^{+}$, and $a b c=1$, prove that
$$\sum b c+2+2 \sqrt{3} \geqslant \frac{5+2 \sqrt{3}}{3} \cdot \sum a$$
Equality holds if and only if $a=b=c=1$. | 54. Simplified proof: Let $\lambda=2+2 \sqrt{3}$, and homogenize the original inequality, i.e., prove
$$\sum b c+\lambda(\sqrt[3]{a b c})^{2} \geqslant \frac{3+\lambda}{3} \cdot \sqrt[3]{a b c} \cdot \sum a$$
Let $a=x^{3}, b=y^{3}, c=z^{3}$, and denote $s_{1}=\sum x=1, s_{2}=\sum y z, s_{3}=x y z$, then inequality (1)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,359 |
55. In $\triangle A B C$, $B C=a, C A=b, A B=c$, and the corresponding exradii are $r_{a}$, $r_{b}, r_{c}$, then
$$\left(\frac{r_{a}}{a}\right)^{2}+\left(\frac{r_{b}}{b}\right)^{2}+\left(\frac{r_{c}}{c}\right)^{2} \geqslant \frac{9}{4}$$
Equality holds if and only if $\triangle A B C$ is an equilateral triangle. | 55. Proof: Let $x=-a+b+c, y=a-b+c, z=a+b-c, x, y, z \in \mathbf{R}^{+}$, then
$$\begin{aligned}
\sum\left(\frac{r_{a}}{a}\right)^{2}-\frac{9}{4}= & {\left[\sum \frac{y z}{x y+z x}-\frac{3}{2}\right]-\sum\left[\frac{1}{4}-\frac{y z}{(y+z)^{2}}\right]=} \\
& \sum \frac{(y-z)^{2}}{2(x+y)(z+x)}-\sum \frac{(y-z)^{2}}{4(y+z)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,360 |
56. (1996 47th Polish Olympiad) Real numbers $a, b, c$ satisfy $\sum a=1, a, b, c \geqslant-\frac{3}{4}$, then
$$\sum \frac{a}{a^{2}+1} \leqslant \frac{9}{10}$$ | 56. Hint: $\frac{a}{a^{2}+1}-\frac{3}{10} \leqslant \frac{18}{25}\left(a-\frac{1}{3}\right) \Leftrightarrow(3 a-1)^{2}(4 a+3) \geqslant 0$.
Generalization: Let real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=1$, and $a_{1}, a_{2}, \cdots, a_{n} \geqslant -\frac{2 n}{n^{2}-1}(n>1)$, then
$$... | \sum \frac{a_{1}}{a_{1}^{2}+1} \leqslant \frac{n^{2}}{n^{2}+1} | Inequalities | proof | Yes | Yes | inequalities | false | 734,361 |
57. Let $x, y, z \in \mathbf{R}^{+}$, then
$$\frac{x^{2}}{y}+\frac{y^{2}}{z}+\frac{z^{2}}{x} \geqslant \sqrt[4]{27 \sum x^{4}}$$
Equality holds if and only if $x=y=z$. | 57. Simplify
$$\begin{array}{l}
\left(\frac{x^{2}}{y}+\frac{y^{2}}{z}+\frac{z^{2}}{x}\right)^{4}-27 \sum x^{4}=\left[\sum x+\sum \frac{(x-y)^{2}}{y}\right]^{4}-27 \sum x^{4}= \\
\left(\sum x\right)^{4}+4\left(\sum x\right)^{3} \cdot \sum \frac{(x-y)^{2}}{y}+6\left(\sum x\right)^{2} \cdot\left[\sum \frac{(x-y)^{2}}{y}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,362 |
Example 25 Let $a, b, c, d \in \overline{\mathbf{R}^{-}}$, then
$$\left(\sum a\right)^{3} \geqslant 4\left[a(c+d)^{2}+b(d+a)^{2}+c(a+b)^{2}+d(b+c)^{2}\right]$$
Equality in (39) holds if and only if $a=c, b=d$. | $$\begin{aligned}
\text { Eq. (39) } \text { RHS }= & 4(a+c)(a+d)(b+c)+4(b+d)(a+b)(c+d) \leqslant \\
& (a+c) \cdot\left(\sum a\right)^{2}+(b+d) \cdot\left(\sum a\right)^{2}= \\
& \left(\sum a\right)^{3}
\end{aligned}$$
Thus, we obtain Eq. (39). The equality condition can easily be seen from the proof process. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,365 |
63. (Original problem, 2007.02.21) $\triangle A B C$ is a non-isosceles triangle, with side lengths $B C=a$, $C A=b$, and $A B=c$, then
$$\left|\frac{b+c}{b-c}+\frac{c+a}{c-a}+\frac{a+b}{a-b}\right|>\frac{17}{3}$$ | 63. Proof: After removing the denominators, the original expression can be transformed into
$$a(b-c)^{2}+b(c-a)^{2}+c(a-b)^{2}>\frac{17}{3}|(a-b)(b-c)(a-c)|$$
Substitute: $a=\frac{y+z}{2}, b=\frac{z+x}{2}, c=\frac{x+y}{2}, x, y, z \in \overline{\mathbf{R}^{-}}$, then equation (1) becomes
$$\begin{array}{l}
(y+z)(y-z)^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,368 |
65. (48th International Mathematical Olympiad China National Training Team Test Question, March 2007) Let positive real numbers $u, v, w$ satisfy $u+v+w+\sqrt{u v w}=4$. Prove that
$$\sqrt{\frac{v w}{u}}+\sqrt{\frac{u w}{v}}+\sqrt{\frac{u v}{w}} \geqslant u+v+w$$ | 65. Proof: Let $\sqrt{\frac{v w}{u}}=x, \sqrt{\frac{u w}{v}}=y, \sqrt{\frac{u v}{w}}=z$, then the original proposition is equivalent to: $x, y, z \in \mathbf{R}^{+}$, and $y z+z x+x y+x y z=4$, then
$$x+y+z \geqslant y z+z x+x y$$
The proof of equation (1) can be found in Exercise 64 (1) of this chapter.
Attachment: T... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,370 |
66. (Original problem, 2006.08.25) Let $x, y, z \in \mathbf{R}^{-}$, and $x^{2}+y^{2}+z^{2}+2 x y z \leqslant 1$, then
$$1+4 x y z \geqslant 2 \sum y z$$
Equality holds if and only if $x=y=z=\frac{1}{2}$, or one of $x, y, z$ is zero, and the other two are equal to $\frac{\sqrt{2}}{2}$. | 66. Proof: Among $2 x-1, 2 y-1, 2 z-1$, there must be two numbers that are both not greater than zero or both not less than zero. Without loss of generality, assume these two numbers are $2 x-1, 2 y-1$, then
$$(2 x-1)(2 y-1) \geqslant 0$$
From this, we get
$$z+4 x y z \geqslant 2 x z+2 y z$$
Additionally, since
$$1 \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,371 |
58. Let $a, b, c \in \mathbf{C}$, then
$$\sum|a|-\sum|b+c|+\left|\sum a\right| \geqslant 0$$ | 58. Simplified Proof
Original expression $\Leftrightarrow\left(\sum|a|+\left|\sum a\right|\right)^{2} \geqslant\left(\sum|b+c|\right)^{2} \Leftrightarrow$
However,
$$\sum(|b| \cdot|c|)+\sum\left(|a| \cdot\left|\sum a\right|\right) \geqslant \sum(|a+b| \cdot|a+c|)$$
$$\begin{array}{l}
\sum(|b| \cdot|c|)+\sum\left(|a| ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,372 |
Example 26 Let $a, b, c \in \mathbf{R}^{+}$, then
$$\frac{a \sqrt{a^{2}+3 b c}}{b+c}+\frac{b \sqrt{b^{2}+3 c a}}{c+a}+\frac{c \sqrt{c^{2}+3 a b}}{a+b} \geqslant a+b+c$$ | To prove that
$$\begin{aligned}
\frac{a \sqrt{a^{2}+3 b c}}{b+c}-a= & \frac{a\left(a^{2}+3 b c\right)}{\sqrt{(b+c)^{2}\left(a^{2}+3 b c\right)}}-a \geqslant \\
& \frac{2 a\left(a^{2}+3 b c\right)}{(b+c)^{2}+\left(a^{2}+3 b c\right)}-a= \\
& \frac{a^{3}+a b c-a\left(b^{2}+c^{2}\right)}{\left(a^{2}+b^{2}+c^{2}\right)+5 b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,376 |
70. (Original problem, 2006. 08.31) Let $a, b, c \in \mathbf{R}^{-}$, then
(1) $a^{3} b+b^{3} c+c^{3} a \leqslant a b c \sum a+\frac{2}{3} \sum b c \cdot \sum(b-c)^{2}$
or $\quad a^{3} b+b^{3} c+c^{3} a \leqslant a b c \sum a+\frac{4}{3}\left(\sum a\right)^{2} \cdot \sum b c-4\left(\sum b c\right)^{2}$
with equality i... | 69. Proof: Since
$$\begin{array}{l}
\sum \frac{(u b+v c)^{2}}{a}= \\
\sum\left\{-(u+v)^{2} a+2(u+v)(u b+v c)+\frac{[(u+v) a-u b-v c]^{2}}{a}\right\}= \\
\quad(u+v)^{2} \sum a+\sum \frac{[(u+v) a-u b-v c]^{2}}{a}
\end{array}$$
Therefore,
$$\begin{array}{l}
{\left[\sum \frac{(u b+v c)^{2}}{a}\right]^{2}-3(u+v)^{4} \sum ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,377 |
73. Let $a, b, c \in \mathbf{R}^{+}$, then
$$\sum \frac{a}{b+c} \geqslant \frac{3}{2}+\frac{\sum(b-c)^{2}}{\left(\sum a\right)^{2}}$$
Equality holds if and only if $a=b=c$. | 73. By symmetry, without loss of generality, assume $a \geqslant b \geqslant c>0$, then
$$\begin{aligned}
\sum \frac{a}{b+c}-\frac{3}{2}-\frac{\sum(b-c)^{2}}{\left(\sum a\right)^{2}}= & \frac{2 \sum a^{3}-\sum a^{2}(b+c)}{2 \prod(b+z)}-\frac{\sum(b-c)^{2}}{\left(\sum a\right)^{2}}= \\
& \frac{\sum(b+c)(b-c)^{2}}{2 \pro... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,380 |
76. (1) In any $\triangle A B C$, we have
$$\sum \frac{\cos ^{2} A}{1+\cos A} \geqslant \frac{1}{2}$$
with equality if and only if $\triangle A B C$ is an equilateral triangle.
(2) In a non-obtuse $\triangle A B C$, we have
$$\sum \frac{\cos ^{2} A}{1+\cos A} \leqslant 2-\sqrt{2}$$
with equality if and only if $\tria... | 76. Proof can be found in "Research on Middle School Mathematics" (Guangdong), Issue 12, 2005, by Yang Xuezhi:
" $\triangle A B C$ with cyclic sum $\sum \frac{\cos ^{2} A}{1+\cos A}$ and its extremum". The problem (1) in question 76 can be used to solve the second problem of the 2005 National High School Mathematics Co... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,381 |
77. (Original problem, 1994. 10.22) Let $D, E, F$ be points on the sides $BC, CA, AB$ of $\triangle ABC$, respectively. Denote the areas of $\triangle EAF, \triangle FBD, \triangle DCE, \triangle DEF$ as $\Delta_{1}, \Delta_{2}, \Delta_{3}, \Delta_{0}, \lambda_{1}, \lambda_{2}, \lambda_{3}$, respectively. Then,
$$\frac... | 77. Proof: Let $\frac{B D}{B C}=x_{1}, \frac{C E}{C A}=x_{2}, \frac{A F}{A B}=x_{3}$, then $\Delta_{1}=\left(1-x_{2}\right) x_{3} \Delta, \Delta_{2}=(1-$ $\left.x_{3}\right) x_{1} \Delta, \Delta_{3}=\left(1-x_{1}\right) x_{2} \Delta\left(\Delta\right.$ is the area of $\triangle A B C$), and we have $\Delta_{0}=\left[x_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,382 |
78. (Proposed by Teacher Wei Liejiao from Jingzhou High School, Hubei Province) If $a, b, c$ are positive numbers, and $a+b+c=1$, then
(1) $\left(\frac{1}{b+c}-a\right)\left(\frac{1}{c+a}-b\right)\left(\frac{1}{a+b}-c\right) \geqslant\left(\frac{7}{6}\right)^{3}$;
(2) $\left(\frac{1}{b+c}+a\right)\left(\frac{1}{c+a}+b\... | 78. Proof: (1) The proof can be found in Wei Liebin's article: "A Pair of Sister Flowers in Inequalities", "Mathematical Communications", Issue 5, 2007.
(2) The condition \(a+b+c=1\) can be relaxed to \(a+b+c \leqslant 1\). Since \(a, b, c \in \overline{\mathbf{R}^{-}}\) and \(a+b+c \leqslant 1\), at least one of \(a, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,383 |
79. (Original problem, 2006.08.18) Let $a, b, c \in \mathbf{R}^{-}$, and $a b c=1$, then
$$2(b+c)(c+a)(a+b) \geqslant 9(a+b+c)-11$$
Equality holds if and only if $a=b=c=1$. | 79. Simplified Proof
Original expression $\Leftrightarrow 2 \sum a \cdot \sum b c-9 \sum a+9 \geqslant 0$
By $\sum b c \geqslant \sqrt{3 a b c \sum a}=\sqrt{3 \sum a}$, to prove (1), it suffices to prove
$$\begin{array}{l}
2 \sum a \cdot \sqrt{3 \sum a}-9 \sum a+9 \geqslant 0 \Leftrightarrow \\
\left(\sqrt{3 \sum a}-3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,384 |
80. (40th IMO Problem) Let $n$ be a fixed integer, $n \geqslant 2$.
(1) Determine the smallest constant $c$ such that the inequality
$$\sum_{1 \leqslant i<j<n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leqslant c\left(\sum_{1 \leqslant i<n} x_{i}\right)^{4}$$
holds for all non-negative numbers $x_{1}, \cdots, x_{n}... | 80. Solution: (Using the local adjustment method) Since the inequality is homogeneous and symmetric, let $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}>0$, and $\sum_{i=1}^{n} x_{i}=1$. At this point, we only need to discuss
$$F\left(x_{1}, \cdots, x_{n}\right)=\sum_{1 \leqslant i<i \leqslant n} x_{i} x_{j}\le... | \frac{1}{8} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,385 |
81. (Original problem, 2007.05.13) Let $a, b, c \in \mathbf{R}$, and $\sum a^{2}=1$, then
$$\sum \sqrt{1-b c} \geqslant \sqrt{7-\sum b c}$$
Equality holds if and only if $a=b=c$. | 81. Proof: Since
$$\begin{aligned}
2 \sqrt{1-a b} \cdot \sqrt{1-a c}= & 2 \sqrt{\sum a^{2}-a b} \cdot \sqrt{\sum a^{2}-a c}= \\
& \sqrt{\sum a^{2}+c^{2}+(a-b)^{2}} \cdot \sqrt{\sum a^{2}+b^{2}+(a-c)^{2}} \geqslant \\
& \sum a^{2}+b c+(a-b)(a-c)= \\
& 1+b c+(a-b)(a-c)
\end{aligned}$$
Similarly, there are two other ineq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,386 |
82. (Original problem, 2007.04.20) Let $x, y, z \in \mathbf{R}^{-}$, then
$$x y z-\prod(-x+y+z) \geqslant \sqrt{2}\left|\prod(y-z)\right|$$
Equality holds if and only if $x=y=z$, or one of $x, y, z$ is zero and the other two are equal. | 82. Simplify
$$\begin{array}{l}
{\left[x y z-\prod(-x+y+z)\right]^{2}-2(y-z)^{2}(z-x)^{2}(x-y)^{2}=} \\
{\left[\sum x(z-x)(x-y)\right]^{2}-2(y-z)^{2}(z-x)^{2}(x-y)^{2}=} \\
{\left[\sum x(z-x)(x-y)\right]^{2}-2(x-y)(y-z)(z-x) \cdot \sum y z(y-z)=} \\
\sum[x(z-x)(x-y)]^{2} \geqslant 0
\end{array}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,388 |
83. (Provided by Ivan Borsenco, Moldova) Let \(a, b, c \in \mathbf{R}^{-}\), then
(1) \(\left(a^{3}+b^{3}+c^{3}\right)^{2} \geqslant\left(a^{4}+b^{4}+c^{4}\right)(b c+c a+a b)\);
(2) \(9\left(a^{4}+b^{4}+c^{4}\right)^{2} \geqslant\left(a^{5}+b^{5}+c^{5}\right)(a+b+c)^{3}\); equality holds in both inequalities if and on... | 83. Proof
(1) Since $\left(\sum a^{3}\right)^{2}=\sum a^{4} \cdot \sum a^{2}-\sum b^{2} c^{2}(b-c)^{2}$
Therefore,
$$\begin{aligned}
\left(\sum a^{3}\right)^{2}-\sum a^{4} \cdot \sum b c= & \sum a^{4} \cdot\left(\sum a^{2}-\sum b c\right)-\sum b^{2} c^{2}(b-c)^{2}= \\
& \frac{1}{2} \sum\left[\left(b^{2}-c^{2}\right)^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,389 |
84. (Dr. Titu Andreescu USA provided) Let $a, b, c \in \mathbf{R}$, then
$$3 \prod\left(b^{2}-b c+c^{2}\right) \geqslant \sum b^{3} c^{3}$$
Equality holds if and only if $a=b=c$. | 84. Proof: (Provided by Dr. Titu Andreescu, USA) If we denote $b+c=s, bc=p$, then
$$\begin{array}{l}
3\left(b^{2}-b c+c^{2}\right)^{3}-\left(b^{6}+b^{3} c^{3}+c^{6}\right)= \\
3\left(s^{2}-3 p\right)^{3}-\left(s^{6}-6 s^{4} p+9 s^{2} p^{2}-p^{3}\right)= \\
\left(s^{2}-4 p\right)^{2}\left(2 s^{5}-5 p\right) \geqslant 0
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,390 |
87. Let $a, b, c \in \mathbf{R}^{+}$, then
$$\frac{1}{\sum a} \cdot \sum \frac{1}{b+c} \geqslant \frac{1}{\sum b c}+\frac{1}{2 \sum a^{2}}$$
Equality holds if and only if $a=b=c$ or one of $a, b, c$ is zero and the other two are equal. | 87. Proof: Let $s_{1}=\sum a, s_{2}=\sum b c, s_{3}=a b c$, then the original inequality is equivalent to
$$s_{1}^{2} s_{2}^{2}+2 s_{1}^{3} s_{3}-4 s_{2}^{3}-3 s_{1} s_{2} s_{3} \geqslant 0$$
Since
$$s_{1}^{2} s_{2}^{2}+18 s_{1} s_{2} s_{3}-4 s_{2}^{3}-4 s_{1}^{3} s_{3}-27 s_{3}^{2}=[(a-b)(b-c)(c-a)]^{2} \geqslant 0$$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,393 |
88. (Provided by Pham kim kung) Let $a, b, c, d \in \overline{\mathbf{R}^{-}}$, and $a+b+c+d=4$, then
$$16+2 a b c d \geqslant 3(a b+a c+a d+b c+b d+c d)$$
Equality holds if and only if $a=b=c=d=1$, or one of $a, b, c, d$ is zero and the other three are equal to $\frac{4}{3}$. | 88. Proof: Let
$$\begin{array}{c}
s_{1}=a+b+c+d \\
s_{2}=a b+a c+a d+b c+b d+c d \\
s_{3}=b c d+a c d+a b d+a b c \\
s_{4}=a b c d
\end{array}$$
Then the original inequality to prove is
$$s_{1}^{4}+32 s_{4} \geqslant 3 s_{1}^{2} s_{2}$$
Inequality (1) can be derived from the following:
$$\begin{aligned}
s_{1}^{4}-3 s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,394 |
89. (Original problem, 2007.05.27) Let $a, b, c \in \overline{\mathbf{R}^{-}}$, and $a+b+c=3$, then
$$\sum \frac{(1+b)(1+c)}{\left(1+b^{2}\right)\left(1+c^{2}\right)} \leqslant 3$$
Equality holds if and only if $a=b=c=1$. | 89. Simplified Proof
$$\begin{array}{l}
\text { Original } \Leftrightarrow \sum(1+b)(1+c)\left(1+a^{2}\right) \leqslant 3 \prod\left(1+a^{2}\right) \Leftrightarrow \\
12-8 \sum b c+3\left(\sum b c\right)^{2}-18 a b c+3(a b c)^{2} \geqslant 0
\end{array}$$
Let $s_{1}=\sum a, s_{2}=\sum b c, s_{3}=a b c$, noting that $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,395 |
90. (Original Question, 2007.05.28) Let $a, b \in \widetilde{\mathbf{R}^{-}}$, and $\frac{a+b}{2} \leqslant 1$, then
$$\frac{1+a}{1+a^{2}}+\frac{1+b}{1+b^{2}} \leqslant \frac{2\left[1+\frac{a+b}{2}\right]}{1+\left(\frac{a+b}{2}\right)^{2}}$$
Equality holds if and only if $a=b$. | 90. Proof: Let $s=\frac{a+b}{2} \leqslant 1, p=a b \leqslant \frac{a+b}{2} \leqslant 1$, then the original inequality is equivalent to
$$\begin{array}{l}
\frac{2+2 s+4 s^{2}-2 p+s p}{1+4 s^{2}-2 p+p^{2}} \leqslant \frac{2(1+s)}{1+s^{2}} \Leftrightarrow \\
2 s^{4}-3 s^{3}-s^{2}+p-p^{2}+3 s p-s p^{2}-s^{2} p+s^{3} p \leq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,396 |
Example 2 Let $a, b, c \in \mathbf{R}^{+}$, and $\sum b c=1$, prove that
$$\frac{1+b^{2} c^{2}}{(b+c)^{2}}+\frac{1+c^{2} a^{2}}{(c+a)^{2}}+\frac{1+a^{2} b^{2}}{(a+b)^{2}} \geqslant \frac{5}{2}$$
Equality holds in (2) if and only if $a=b=c=\frac{\sqrt{3}}{3}$. | Prove:
$$\begin{aligned}
\sum \frac{1+b^{2} c^{2}}{(b+c)^{2}}-\frac{5}{2}= & \sum \frac{\left(1+b^{2}\right)\left(1+c^{2}\right)+2 b c}{(b+c)^{2}}-\frac{11}{2}= \\
& \sum \frac{(b+c)^{2}(a+b)(a+c)+2 b c}{(b+c)^{2}}-\frac{11}{2}= \\
& \frac{1}{2} \sum(b-c)^{2}-\frac{1}{2}\left(\sum b c\right) \cdot \sum\left(\frac{b-c}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,398 |
92. (Vasile Cartoage Romania) If $x, y, z \in \widetilde{\mathbf{R}^{-}}$, then
$$\sum x^{4}(y+z) \leqslant \frac{1}{12}\left(\sum x\right)^{5}$$
Equality holds if and only if $x=y=z=0$, or one of $x, y, z$ is zero, and the other two are $\frac{1}{2}+\frac{\sqrt{3}}{6}$, $\frac{1}{2}-\frac{\sqrt{3}}{6}$, respectively. | 92. Proof: By homogeneity, we can assume $x+y+z=1$. If $x, y, z$ are all no greater than $\frac{1}{2}$, then
$$\begin{aligned}
\sum x^{4}(y+z)= & \sum x^{4}(1-x)<\frac{1}{8} \sum x(1-x)= \\
& \frac{1}{8} \sum x-\frac{1}{8} \sum x^{2} \leqslant \\
& \frac{1}{8}-\frac{1}{8} \cdot \frac{1}{3}\left(\sum x\right)^{2}=\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,399 |
96. (Original problem, 1986. 03. 20) Let $\beta_{1}, \beta_{2}, \beta_{3} \in [0, \pi], \varphi_{1}, \varphi_{2}, \varphi_{3} \in \mathbf{R}$ and $\sum \beta_{1} = \sum \varphi_{1} = \pi$, then
$$\sum \sin \varphi_{1}\left(-\sin \beta_{1}+\sin \beta_{2}+\sin \beta_{3}\right) \leqslant 2\left(1+\sin \frac{\beta_{1}}{2} ... | 96. Proof
$$\begin{array}{l}
\sum \sin \varphi_{1}\left(-\sin \beta_{1}+\sin \beta_{2}+\sin \beta_{3}\right)= \\
4 \sum \sin \frac{\beta_{2}}{2} \sin \frac{\beta_{3}}{2} \cos \frac{\beta_{1}}{2} \sin \varphi_{1}= \\
4 \cos \frac{\beta_{1}}{2} \cos \frac{\beta_{2}}{2} \cos \frac{\beta_{3}}{2} \sum \tan \frac{\beta_{2}}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,403 |
97. (Provided by Pham Kim Hung) Let $a, b, c$ be non-negative numbers, and $a+b+c=3$, then
$$a^{4}+b^{4}+c^{4}-3 a b c \geqslant 6 \sqrt{2}(a-b)(b-c)(c-a)$$
Equality holds if and only if $a=b=c$; or $a=\frac{3(1-\sqrt{5})}{2}, b=\frac{3(\sqrt{5}-1)}{2}, c=0$; or $b=$ $\frac{3(1-\sqrt{5})}{2}, c=\frac{3(\sqrt{5}-1)}{2}... | 97. Proof: For non-negative numbers $x, y, z, t$, we first examine the function
$$\begin{aligned}
f(t)= & \sum(x+t)^{4}-\Pi(x+t) \cdot \sum(x+t)- \\
& 2 \sqrt{2}(x-y)(y-z)(z-x) \cdot \sum(x+t)= \\
& A+B t+c t^{2}
\end{aligned}$$
That is $\square$
$$\begin{array}{c}
A=\sum x^{4}-x y z \sum x-2 \sqrt{2}(x-y)(y-z)(z-x) \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,404 |
100. (Original problem, 2006.02.12) Let $x_{1}, x_{2}, x_{3}, x_{4} \in \overline{\mathbf{R}^{-}}$, and $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2} \leqslant 3$, then
$$\begin{aligned}
3 x_{1} x_{2} x_{3} x_{4} \geqslant & \sum x_{2} x_{3} x_{4}-1 \geqslant \sum_{1 \leqslant i<j \leqslant 4} x_{i} x_{j}-3 \geqslant \\
& 3... | 100. Proof see "A Chain of Inequalities and Its Proof" by Yang Xuezhi in *Mathematics Teaching in Middle Schools* (Anhui), 2007, Issue 4.
Conjecture: Let $x_{1}, x_{2}, \cdots, x_{n} \in \overline{\mathbf{R}^{-}}$, and $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2} \leqslant n-1$, prove or disprove:
(1) $\frac{1}{2}(n-1)(n-2)+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,407 |
101. (Original problem, 2007. 12.20) Let $a_{i} \in \mathbf{R}^{+}, i=1,2, \cdots, n, m, k$ be positive integers, then
$$\sum_{i=1}^{n} \frac{a_{i}^{m+k}}{a_{i+1}^{m}} \geqslant \sum_{i=1}^{n} \frac{a_{i}^{m}}{a_{i+1}^{m-k}}$$
Equality holds if and only if $a_{1}=a_{2}=\cdots=a_{n}$. | 101. Simplify and prove
$$\begin{aligned}
\sum_{i=1}^{n} \frac{\left(a_{i}^{m}-a_{i+1}^{m}\right)\left(a_{i}^{k}-a_{i+1}^{k}\right)}{a_{i+1}^{m}}= & \sum_{i=1}^{n}\left(\frac{a_{i}^{m+k}}{a_{i+1}^{m}}-\frac{a_{i}^{m}}{a_{i+1}^{m-k}}-a_{i}^{k}+a_{i+1}^{k}\right)= \\
& \sum_{i=1}^{n} \frac{a_{i}^{m+k}}{a_{i+1}^{m}}-\sum_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,408 |
102. (Original problem, 2007. 12. 28) Let $a, b, c \in \mathbf{R}^{+}$, and $\sum b c=1$, then
$$\sum \frac{1}{4 a^{2}-b c+2} \geqslant 1$$
Equality holds if and only if $a=b=c=\frac{\sqrt{3}}{3}$. | 103. Proof: We transform the proposition into an equivalent form by setting $x=2a, y=2b, z=2c, a, b, c \in \overline{\mathbf{R}^{-}}$. Thus, we obtain the following equivalent proposition:
Given $a, b, c \in \overline{\mathbf{R}^{-}}$, and satisfying
$$\begin{array}{c}
2 \sum bc - \sum a^2 = 4abc \\
(1-2a)(1-2b)(1-2c)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,410 |
105. (Original problem, 2007. 12.21) Let $x, y, z \in \mathbf{R}^{-}$, then
$$\left(x^{2}-y^{2}-x y\right)^{2}+\left(y^{2}-z^{2}-y z\right)^{2}+\left(z^{2}-x^{2}-z x\right)^{2} \geqslant y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2}$$
Equality holds if and only if $x=y=z$. | 105. Simplified proof: The original equation is equivalent to
$$\sum\left(x^{2}-y^{2}\right)^{2} \geqslant 2 \sum x y\left(x^{2}-y^{2}\right)$$
When $x \geqslant y \geqslant z$, we have
$$\sum x y\left(x^{2}-y^{2}\right)=(x+y+z)(x-y)(y-z)(x-z) \geqslant 0$$
Therefore, to prove equation (1), it suffices to prove that ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,413 |
107. Let $x, y, z \in \mathbf{R}^{+}$, and $\sum x \geqslant \frac{y z}{x}, \frac{z x}{y}, \frac{x y}{z}$, then
$$\sum x \cdot \sum \frac{1}{(y+z)^{4}} \geqslant \frac{9}{16 x y z}$$
Equality holds if and only if $x=y=z$. | 107. Proof: Let $x=-a+b+c, y=a-b+c, z=a+b-c$, then from the condition $\sum x \geqslant \frac{y z}{x}, \frac{z x}{y}, \frac{x y}{z}$, we can obtain $b^{2}+c^{2} \geqslant a^{2}, c^{2}+a^{2} \geqslant b^{2}, a^{2}+b^{2} \geqslant c^{2}$, therefore, the original problem is equivalent to: In a non-acute $\triangle A B C$,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,415 |
108. (Original problem, 1988.11.10) Let \(a, b, c, d, x, y, z, w \in \mathbf{R}^{+}\), then
$$\begin{array}{c}
x(-a+b+c+d)+y(a-b+c+d)+ \\
z(a+b-c+d)+w(a+b+c-d) \leqslant \\
\sqrt{\frac{(x y+z w)(x z+y w)(x w+y z)}{x y z w}} . \\
\sqrt{\frac{(a b+c d)(a c+b d)(a d+b c)}{a b c d}}
\end{array}$$
Equality holds if and onl... | 108. Proof: Let $M=\sqrt{\frac{(a b+c d)(a c+b d)(a d+b c)}{a b c d}}$, then it is easy to prove that
$$\begin{array}{l}
(-a+b+c d)^{2}+\frac{(-b c d+a c d+a b d+a b c)^{2}}{a b c d}= \\
(a-b+c+d)^{2}+\frac{(b c d-a c d+a b d+a b c)^{2}}{a b c d}= \\
(a+b-c+d)^{2}+\frac{(b c d+a c d-a b d+a b c)^{2}}{a b c d}= \\
(a+b+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,416 |
109. (Chen Ji, 2007. 12. 20 provided) Let $x, y, z \in \mathbf{R}^{-}$, and satisfy $x+y+z=2$, then
$$\sum \sqrt{9-8 y z} \geqslant 7$$
Equality holds if and only if $x=y=z=\frac{2}{3}$, or one of $x, y, z$ is zero, and the other two are equal to 1. | 109. Simplified Proof: For the convenience of proof, we make the following transformation: Let $x=2a, y=2b, z=2c$. At this point, the original proposition is equivalent to:
Let $a, b, c \in \overline{\mathbf{R}^{-}}$, and satisfy $a+b+c=1$, then
$$\sum \sqrt{9-32bc} \geqslant 7$$
Equality holds if and only if $a=b=c=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,417 |
110. (Original problem, 1990.01.14) Let the sides of $\triangle A B C$ be $B C=a, C A=b, A B=c$, and $\angle A \geqslant \frac{\pi}{3}$, then
$$(b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 4+\frac{1}{\sin \frac{A}{2}}$$
Equality holds if and only if $b=c$. | 110. Simplified proof: The original expression is equivalent to
$$\begin{aligned}
\frac{b}{c}+\frac{c}{b}-2 \geqslant & \sqrt{\frac{4 b c}{(a-b+c)(a+b-c)}}-\frac{b+c}{a} \Leftrightarrow \\
& \frac{(b-c)^{2}}{b c}\left[\sqrt{\frac{4 b c}{(a-b+c)(a+b-c)}}+\frac{b+c}{a}\right] \geqslant \\
& \frac{4 b c}{(a-b+c)(a+b-c)}-\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,418 |
Example 4 Let $x, y, z \in \mathbf{R}^{+}$, prove that
$$\sum \frac{y z}{x(y+z)^{2}} \geqslant \frac{9}{4 \cdot \sum x}$$
The equality in (4) holds if and only if $x=y=z$. | Prove
$$\begin{array}{l}
4 \sum x \cdot \sum \frac{y z}{x(y+z)^{2}}-9=\left(4 \sum \frac{y z}{(y+z)^{2}}-3\right)+\left(4 \sum \frac{y z}{x(y+z)}-6\right)= \\
2 \sum \frac{x^{2}(y-z)^{2}}{y z(x+y)(x+z)}-\sum\left(\frac{y-z}{y+z}\right)^{2}= \\
\frac{1}{\prod(y+z)} \cdot \sum \frac{3 x^{2} y z+2 x^{2} y^{2}+2 x^{2} z^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,420 |
112. ("Home of Olympiad Mathematics" website, 2007.08.07 provided) Let $a, b, c \in \mathbf{R}^{+}$, and $a b c=1$, then
$$\sum \sqrt{3 a^{2}+4} \leqslant \sqrt{7} \cdot \sum a$$
Equality holds if and only if $a=b=c=1$. | 112. Proof: We divide the proof into two cases.
(1) When $\sum a \geqslant \sum b c$, we have
$$\begin{aligned}
\sum \sqrt{3 a^{2}+4}= & \sum \sqrt{a(3 a+4 b c)} \leq \\
& \sqrt{\sum a \cdot \sum(3 a+4 b c)}= \\
& \sqrt{\sum a \cdot\left(3 \sum a+4 \sum b c\right)} \leq \\
& \sqrt{7} \sum a
\end{aligned}$$
The origina... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,421 |
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