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Example 2 Let $\frac{3}{2} \leqslant x \leqslant 5$, prove the inequality: $2 \sqrt{x+1}$ $+\sqrt{2 x-3}+\sqrt{15-3 x}<2 \sqrt{19}$. (2003 National Mathematics League Question) | Prove: $2 \sqrt{x+1}+\sqrt{2 x-3}+\sqrt{15-3 x}=$ $\sqrt{2} \times \sqrt{2 x+2}+1 \times \sqrt{2 x-3}+1 \times \sqrt{15-3 x} \leqslant$ $\sqrt{\left(\sqrt{2}^{2}+1^{2}+1^{2}\right)\left[(\sqrt{2 x+2})^{2}+(\sqrt{2 x-3})^{2}+(\sqrt{15-3 x})^{2}\right]}$ $=2 \sqrt{x+14} \leqslant 2 \sqrt{19}$,
$\because \frac{\sqrt{2 x+2... | 2 \sqrt{x+1}+\sqrt{2 x-3}+\sqrt{15-3 x} < 2 \sqrt{19} | Inequalities | proof | Yes | Yes | inequalities | false | 736,975 |
Example 3 Use the Cauchy inequality to prove Pedoe's inequality: Let the three sides of two triangles be $a, b, c$ and $x, y, z$, and their areas be $\Delta_{1}, \Delta_{2}$, then there is the inequality $\left(a^{2} + b^{2} + c^{2}\right)\left(x^{2} + y^{2} + z^{2}\right) \geqslant 2\left(a^{2} x^{2} + b^{2} y^{2} + c... | Proof: From Heron's formula $\triangle=$
$$\begin{array}{l}
\sqrt{s(s-a)(s-b)(s-c)}\left(s=\frac{a+b+c}{2}\right) \text {, we get: } 16 \\
\Delta^{2}=2 s(2 s-2 a)(2 s-2 b)(2 s-2 c)=\left(a^{2}+\right. \\
\left.b^{2}+c^{2}\right)^{2}-2\left(a^{4}+b^{4}+c^{4}\right) . \\
\quad \therefore 16 \Delta_{1} \Delta_{2}+2\left(a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,976 |
Example 4 Let $P$ be a point inside $\triangle ABC$, and draw perpendiculars from $P$ to the three sides $BC$, $CA$, and $AB$, with the feet of the perpendiculars being $D$, $E$, and $F$ respectively. For what position of $P$ is the value of $\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$ minimized? (22nd IMO problem) | Solution: Let $B C=a, C A=b$, $A B=c, P D=d_{1}, P E=d_{2}$, $P F=d_{3},$ and the area of $\triangle A B C$ be $\Delta$. Clearly, we have $2 \Delta=a d_{1}+b d_{2}+c d_{3}$.
$$\begin{array}{c}
\therefore \frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F}=\frac{a}{d_{1}}+ \\
\frac{b}{d_{2}}+\frac{c}{d_{3}} \geqslant \frac{... | \frac{(a+b+c)^{2}}{2 \Delta} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,977 |
Example 6 Given $a_{1}, a_{2}, \cdots, a_{n} \in R^{+}$ and $a_{1}+a_{2}+\cdots+a_{n}=1$, prove: $\frac{a_{1}^{2}}{a_{1}+a_{2}}+\frac{a_{2}^{2}}{a_{2}+a_{3}}+\cdots+$ $\frac{a_{n}^{2}}{a_{n}+a_{1}} \geqslant \frac{1}{2}$ (Problem from the 24th Soviet Union Mathematical Competition). | Proof: From the transformation 3 of the Cauchy inequality, we get:
\[
\frac{a_{1}^{2}}{a_{1}+a_{2}}+
\frac{a_{2}^{2}}{a_{2}+a_{3}}+\cdots+\frac{a_{n}^{2}}{a_{n}+a_{1}} \geqslant \frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}}{2\left(a_{1}+a_{2}+\cdots+a_{n}\right)}= \\
\frac{1}{2}
\] | \frac{1}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,978 |
For example, $6 n \in N_{+}$ and $n \geqslant 2$, prove $\frac{4}{7}<1-\frac{1}{2}+$
$$\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2 n-1}-\frac{1}{2 n}<\frac{\sqrt{2}}{2}$$ | $$\begin{array}{l}
\text { Proof: } \because 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2 n-1}- \\
\frac{1}{2 n}=1-\left(1-\frac{1}{2}\right)+\frac{1}{3}-\left(\frac{1}{2}-\frac{1}{4}\right)+\cdots+ \\
\frac{1}{2 n-1}-\left(\frac{1}{n}-\frac{1}{2 n}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,979 |
Let $a_{i}, x_{i}, y_{i}(i=1,2, \cdots, n)$ be any real numbers, then
(i) when $p>1$,
$$\begin{array}{l}
\quad\left[\sum_{i=1}^{n}\left|a_{i}\right|\left|x_{i}+y_{i}\right|^{p}\right]^{\frac{1}{p}} \leqslant\left(\sum_{i=1}^{n}\left|a_{i}\right|\left|x_{i}\right|^{p}\right)^{\frac{1}{p}}+ \\
\left(\sum_{i=1}^{n}\left|a... | $$\begin{array}{l}
\text { Prove: (i) When } p>1 \text {, } \\
{\left[\sum_{i=1}^{n}\left|a_{i}\right|\left|x_{i}+y_{i}\right|^{p}\right]^{\frac{1}{p}}} \\
=\left[\sum_{i=1}^{n}\left|a_{i}{ }^{\frac{1}{p}} x_{i}+a_{i}{ }^{\frac{1}{p}} y_{i}\right|^{p}\right]^{\frac{1}{p}}, \\
\left.\sum_{i=1}^{n}\left|a_{i}^{\frac{1}{p... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,981 |
Let $x(t), y(t)$ be measurable functions on the measurable space $\left(T, \sum, \mu\right)$, then the following inequalities hold:
$$\begin{array}{l}
\text { (i) When } p>1 \text {, } \\
{\left[\int_{T}|x(t)+y(t)|^{p} d \mu\right]^{\frac{1}{p}}} \\
\leqslant\left[\int_{T}|x(t)|^{p} d \mu\right]^{\frac{1}{p}}+\left[\in... | Proof: (i) When $p>1$
Let $x(t), y(t) \in L^{p}(T, \mu)$, then $x(t)+y(t) \in L^{p}(T, \mu)$. Take $q>1$, such that $\frac{1}{p}+\frac{1}{q}=1$, then $|x(t)+y(t)|^{p / q} \in L^{p}(T, \mu)$. By the $H \ddot{o} l d e r$ inequality, we get:
$$\begin{array}{l}
\int_{T}|x(t)||x(t)+y(t)|^{p / q} d \mu \\
\leqslant\left(\int... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,982 |
Let $P(a, b)$ be a point in the first quadrant of the Cartesian coordinate system. Draw a line through point $P$ that intersects the two positive semi-axes at points $A$ and $B$. When the length of $AB$ is minimized, find the intercepts of $AB$ on the $x$-axis and $y$-axis. | Solution: As shown in the figure:
Let the intercepts of line $AB$ on the $x$-axis and $y$-axis be $s$ and $t$, respectively, then we have $\frac{a}{s}+\frac{b}{t}=1$ (1), and the length of line segment $AB$ is: $l=\left(s^{2}+t^{2}\right)^{\frac{1}{2}}$
By Hölder's inequality $\left(p=3, q=\frac{3}{2}\right)$ and cond... | s=a+a^{\frac{1}{3}} b^{\frac{2}{3}}, t=b+a^{\frac{2}{3}} b^{\frac{1}{3}} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,985 |
If $a>0, b>0, p>1, q>1$ and $\frac{1}{p}+\frac{1}{q}=1$ then $a b \leqslant \frac{1}{p} a^{p}+\frac{1}{q} b^{q}$ | Prove by Lemma $1 . x>0 \quad x^{a}-\alpha x \leqslant 1-\alpha(01$
and $\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=\frac{1}{\mathrm{p}}+\frac{\mathrm{p}-1}{\mathrm{p}}=1$
Secondly, for $x^{a} \leqslant \alpha x+1-\alpha \quad(x>0,00, b>0, \frac{a^{p}}{b^{q}}>0)$ we get $\left(\frac{a^{p}}{b^{q}}\right)^{\frac{1}{p}} \l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,997 |
Let $x_{i}>0, y_{i}>0(i=1,2, \cdots n)$ be a finite number of positive numbers, and $p>1, q>1 \quad \frac{1}{p}+\frac{1}{q}=1 \quad$ prove $\sum_{i=1}^{n} x_{i} y_{i} \leqslant\left(\sum_{i=1}^{n} x_{i}^{p}\right)^{\frac{1}{p}}\left(\sum_{i=1}^{n} y_{i}^{q}\right)^{\frac{1}{q}}(H$ ölder inequality $(I))$ | Proof: By Lemma 2, we have
$$a b \leqslant \frac{1}{p} a^{p}+\frac{1}{q} b^{q}$$
where $a>0, b>0; p>1, q>1$ and $\frac{1}{p}+\frac{1}{q}=1$. Making the substitution, let $a^{p}=A$ and $b^{q}=B$, then we have
$$A \frac{1}{p} B^{\frac{1}{q}} \leqslant \frac{1}{p} A+\frac{1}{q} B$$
Furthermore, by the given conditions, $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,999 |
Let $f(x)$ and $g(x)$ be continuous on $[a, b]$, $p>1, q>1$ and $\frac{1}{p}+\frac{1}{q}=1$ prove $\int_{a}^{b}|f(x) g(x)| d x \leqslant\left(\int_{a}^{b}|f(x)|^{p} d x\right)^{\frac{1}{p}}\left(\int_{a}^{b}|g(x)|^{q} d x\right)^{\frac{1}{q}}$
(Hölder's Inequality (II)) | Proof: By Hölder's inequality (I), we know that
$$\sum_{i=1}^{n}\left|f_{i} g_{i}\right| \leqslant\left(\sum_{i=1}^{n}\left|f_{i}\right|^{p}\right)^{\frac{1}{p}}\left(\sum_{i=1}^{n}\left|g_{i}\right|^{q}\right)^{\frac{1}{q}}$$
Here, $\quad f_{i}=f\left(x_{i}\right), g_{i}=g\left(x_{i}\right)$
$$\begin{array}{rr}
x_{i}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,000 |
Let $x_{i}>0, y_{i}>0 \quad(i=1,2, \cdots n)$ be a finite number of positive numbers, and $\lambda>1$. Prove that
$$\left(\sum_{i=1}^{n}\left(x_{i}+y_{i}\right)^{\lambda}\right)^{\frac{1}{\lambda}} \leqslant\left(\sum_{i=1}^{n} x_{i}^{\lambda}\right)^{\frac{1}{\lambda}}+\left(\sum_{i=1}^{n} y_{i}^{\lambda}\right)^{\fra... | Let $p=\lambda, q=\frac{\lambda}{\lambda-1}$, then $p>1, q>1$, and we have $\frac{1}{p}+\frac{1}{q}=\frac{1}{\lambda}+\frac{\lambda-1}{\lambda}=1$. Thus,
$$\sum_{i=1}^{n}\left(x_{i}+y_{i}\right)^{\lambda}=\sum_{i=1}^{n} x_{i}\left(x_{i}+y_{i}\right)^{\lambda-1}+\sum_{i=1}^{n} y_{i}\left(x_{i}+y_{i}\right)^{\lambda-1}$$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,001 |
Example 1 Given $a \sqrt{1-b^{2}}+b \sqrt{1-a^{2}}=1$, prove:
$$a^{2}+b^{2}=1$$ | By the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& \left(a \sqrt{1-b^{2}}+b \sqrt{1-a^{2}}\right)^{2} \\
\leqslant & {\left[a^{2}+\left(\sqrt{1-a^{2}}\right)^{2}\right]\left[\left(\sqrt{1-b^{2}}\right)^{2}+b^{2}\right]=1 }
\end{aligned}$$
The equality holds if and only if there exists a constant $k$ such th... | a^{2}+b^{2}=1 | Algebra | proof | Yes | Yes | inequalities | false | 737,003 |
Example 2 Given $\frac{\cos ^{4} A}{\cos ^{2} B}+\frac{\sin ^{4} A}{\sin ^{2} B}=1$, find the value of $\frac{\cos ^{4} B}{\cos ^{2} A}+\frac{\sin ^{4} B}{\sin ^{2} A}$. | By Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& \frac{\cos ^{4} A}{\cos ^{2} B}+\frac{\sin ^{4} A}{\sin ^{2} B} \\
= & {\left[\left(\frac{\cos ^{2} A}{\cos B}\right)^{2}+\left(\frac{\sin ^{2} A}{\sin B}\right)^{2}\right]\left[\cos ^{2} B+\sin ^{2} B\right] } \\
\geqslant & {\left[\frac{\cos ^{2} A}{\cos B} \c... | 1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,004 |
Example 3 Given the relationships between real numbers $m, n, a, b$ and angle $\theta$: $m \sin \theta - n \cos \theta = \sqrt{m^{2} + n^{2}}, \frac{\sin ^{2} \theta}{a^{2}} + \frac{\cos ^{2} \theta}{b^{2}} = 1$, prove that: $\frac{m^{2}}{a^{2}} + \frac{n^{2}}{b^{2}} = m^{2} + n^{2}$. | Prove that by the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
(m \sin \theta - n \cos \theta)^{2} & \leqslant \left(m^{2} + n^{2}\right)\left(\sin ^{2} \theta + \cos ^{2} \theta\right) \\
& = m^{2} + n^{2}
\end{aligned}$$
The equality holds if and only if there exists a constant \( k \) such that
$$\left\{\be... | \frac{m^{2}}{a^{2}} + \frac{n^{2}}{b^{2}} = m^{2} + n^{2} | Algebra | proof | Yes | Yes | inequalities | false | 737,005 |
Example 4 Solve the equation $\sqrt{12-\frac{12}{x^{2}}}+\sqrt{x^{2}-\frac{12}{x^{2}}}=x^{2}$. | Solve the original equation, which can be rewritten as $\frac{\sqrt{12}}{x^{2}} \cdot \sqrt{1-\frac{1}{x^{2}}}+\frac{1}{|x|}$. $\sqrt{1-\frac{12}{x^{4}}}=1$, and by the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& \left(\frac{\sqrt{12}}{x^{2}} \cdot \sqrt{1-\frac{1}{x^{2}}}+\frac{1}{|x|} \cdot \sqrt{1-\frac{1... | x = \pm 2 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,006 |
Example 1 Suppose $a^{2}+b^{2}+c^{2}=4, x^{2}+y^{2}+z^{2}=9$, then the range of values for $a x+b y+c z$ is $\qquad$ (Example 9 from [1]) | By Cauchy-Schwarz inequality, we have $4 \times 9=\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}\right.$ $\left.+y^{2}+z^{2}\right) \geqslant(a x+b y+c z)^{2}, \therefore-6 \leqslant a x+b y+c z$ $\leqslant 6$ | -6 \leqslant a x+b y+c z \leqslant 6 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,007 |
Example 2 Given that $a, b, c, d, e$ are real numbers satisfying $a+b+c+d$ $+e=8$ and $a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16$, find the range of values for $e$. (Example 1 from [1]) | By Cauchy-Schwarz inequality, we have $\left(1^{2}+1^{2}+1^{2}+1^{2}\right)\left(a^{2}+b^{2}+c^{2}+d^{2}\right) \geqslant(a+b+c+d)^{2}, \therefore 4\left(16-e^{2}\right) \geqslant(8-e)^{2}, \therefore 0 \leqslant e \leqslant \frac{16}{5}$.
Using the same approach as in Example 2 of this article, we can solve Example 2... | 0 \leqslant e \leqslant \frac{16}{5} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,008 |
Example 3 Given $x+2 y+3 z+4 u+5 v=30$, find the minimum value of $w$ $=x^{2}+2 y^{2}+3 z^{2}+4 u^{2}+5 v^{2}$ (Example 6 from reference [1]). | By Cauchy-Schwarz inequality, we have $(1+2+3+4+5)\left(x^{2}+\right.$ $\left.2 y^{2}+3 z^{2}+4 u^{2}+5 v^{2}\right) \geqslant(x+2 y+3 z+4 u+5 v)^{2}$, i.e., $15 w \geqslant 30^{2}, \therefore$ the minimum value of $w$ is 60. (At this point, $x=y=z$ $=u=v=2)$ | 60 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,009 |
Example 4 Given real numbers $x_{1}, x_{2}, x_{3}$ satisfy $x_{1}+\frac{1}{2} x_{2}+$ $\frac{1}{3} x_{3}=1$ and $x_{1}^{2}+\frac{1}{2} x_{2}^{2}+\frac{1}{3} x_{3}^{2}=3$, then the minimum value of $x_{3}$ is $\qquad$ . (Example 7 in [1]) | By Cauchy-Schwarz inequality, we have $\left(1+\frac{1}{2}\right)\left(x_{1}^{2}+\frac{1}{2} x_{2}^{2}\right) \geqslant$ $\left(x_{1}+\frac{1}{2} x_{2}\right)^{2}$, i.e., $\frac{3}{2}\left(3-\frac{1}{3} x_{3}^{2}\right) \geqslant\left(1-\frac{1}{3} x_{3}\right)^{2}, \therefore$ $-\frac{21}{11} \leqslant x_{3} \leqslant... | -\frac{21}{11} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,010 |
Example 5 Find the values of real numbers $x, y$ such that $(y-1)^{2}+(x+$ $y-3)^{2}+(2 x+y-6)^{2}$ reaches the minimum value. (Example 8 in [1]) | $$\begin{array}{l}
\text { To solve, introduce parameters } u, v, w, \text { by Cauchy-Schwarz inequality we have } \\
\left(u^{2}+v^{2}+w^{2}\right)\left[(y-1)^{2}+(x+y-3)^{2}+(2 x\right. \\
\left.+y-6)^{2}\right] \geqslant[u(y-1)+v(x+y-3)+w(2 x+y \\
-6)]^{2}=[(v+2 w) x+(u+v+w) y-u-3 v- \\
6 w]^{2},
\end{array}$$
To ... | x=\frac{5}{2}, y=\frac{5}{6} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,011 |
Example 6 Solve the system of equations $\left\{\begin{array}{l}x+y+z=3, \\ x^{2}+y^{2}+z^{2}=3,(\text{Example 4 in [1]}) \\ x^{5}+y^{5}+z^{5}=3 .\end{array}\right.$ | Solve: By Cauchy-Schwarz inequality, we have $\left(1^{2}+1^{2}+1^{2}\right)\left(x^{2}+y^{2}+\right.$ $\left.z^{2}\right) \geqslant(x+y+z)^{2}$, and $x+y+z=3, x^{2}+y^{2}+z^{2}=$ 3, so the equality of the above inequality holds, thus $x=y=z=1$, which also satisfies $x^{5}+y^{5}+z^{5}=3$, hence the solution to the orig... | x=y=z=1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,012 |
Example 7 Find all real number pairs $(x, y)$ that satisfy the equation $x^{2}+(y-1)^{2}+(x-y)^{2}=$ $\frac{1}{3}$. (Example 5 from [1]) | Introducing undetermined parameters $u, v, w$, by the Cauchy-Schwarz inequality we have:
\[
\begin{array}{l}
\left(u^{2}+v^{2}+w^{2}\right)\left[x^{2}+(y-1)^{2}+(x-y)^{2}\right] \geqslant \\
{[u x+v(y-1)+w(x-y)]^{2} }=[(u+w) x+(v-
[v]^{2} }=[(u+w) x+(v-w) y-v]^{2}
\end{array}
\]
Let $u+w=0, v-w=0$, we can take $u=-1, ... | x=\frac{1}{3}, y=\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,013 |
Example 8 If $a+b+c=1$, then the maximum value of $\sqrt{3 a+1}+$ $\sqrt{3 b+1}+\sqrt{3 c+1}$ is $\qquad$ . (Example 10 from [1]) | By Cauchy-Schwarz inequality, we have $(\sqrt{3 a+1}+\sqrt{3 b+1}+$ $\sqrt{3 c+1})^{2} \leqslant\left(1^{2}+1^{2}+1^{2}\right)[(3 a+1)+(3 b+1)+$ $(3 c+1)]$,
Combining $a+b+c=1$, we get $\sqrt{3 a+1}+\sqrt{3 b+1}+$ $\sqrt{3 c+1} \leqslant 3 \sqrt{2}$, so the maximum value we seek is $3 \sqrt{2}$. (At this point, $a=$ $... | 3 \sqrt{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,014 |
Example 9 The range of the function $y=\sqrt{1994-x}+\sqrt{x-1993}$ is __. (Example 11 in [1]) | By Cauchy-Schwarz inequality, we have $(\sqrt{1994-x}+$
$$\begin{array}{l}
\sqrt{x-1993})^{2} \leqslant\left(1^{2}+1^{2}\right)[(1994-x)+(x-1993)] \\
=2, \therefore y \leqslant \sqrt{2} .
\end{array}$$
On the other hand, $y^{2}=1+2 \sqrt{1994-x} \cdot \sqrt{x-1993} \geqslant$ $1, \therefore y \geqslant 1$, hence the r... | [1, \sqrt{2}] | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,015 |
Example 3 Let $x, y, z$ be non-negative numbers, and $x^{2}+y^{2}+z^{2}=3$ Prove: $\frac{x}{\sqrt{x^{2}+y+z}}+\frac{y}{\sqrt{y^{2}+z+x}}+\frac{z}{\sqrt{z^{2}+x+y}}$ $\leqslant \sqrt{3} \quad$ This problem is still manageable | Prove: By the local Cauchy inequality, we have: $\left(x^{2}+y+z\right)$ $(1+y+z) \geqslant(x+y+z)^{2}$. Therefore, $\frac{1}{\sqrt{x^{2}+y+z}} \leqslant$ $\frac{\sqrt{1+y+z}}{x+y+z}$, and similarly, we can obtain the other two inequalities. Also, by the Cauchy inequality $3\left(x^{2}+y^{2}+z^{2}\right) \geqslant(x+y+... | \sqrt{3} | Inequalities | proof | Yes | Yes | inequalities | false | 737,017 |
Proposition 2 If the series $\sum_{n=1}^{\infty} a_{n}^{2}$ and $\sum_{n=1}^{\infty} b_{n}^{2}$ converge, then the inequality
$$\left(\sum_{n=1}^{\infty} a_{n} b_{n}\right)^{2} \leq \sum_{n=1}^{\infty} a_{n}^{2} \sum_{n=1}^{\infty} b_{n}^{2}$$
holds. | Proof: $\because \sum_{n=1}^{\infty} a_{n}^{2} \sum_{n=1}^{\infty} b_{n}^{2}$ converges,
$$0 \leq\left(\sum_{i=1}^{\infty} a_{i} b_{i}\right)^{2} \leq\left(\sum_{i=1}^{\infty} a_{i}^{2}\right)\left(\sum_{i=1}^{\infty} b_{i}^{2}\right)$$
$\therefore \sum_{n=1}^{\infty} a_{n} b_{n}$ converges, and
$$\left(\lim _{n \right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,027 |
Proposition 3 If the series $\sum_{i=1}^{\infty} a_{1}^{2}$ and $\sum_{i=1}^{\infty} b_{i}^{2}$ converge, and for $\forall n \in N$ we have
$$\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{2} \leq\left(\sum_{i=1}^{n} a_{i}^{2}\right)\left(\sum_{i=1}^{n} b_{i}^{2}\right)$$
then for any continuous functions $f(x), g(x)$ defin... | Proof: Since the functions $f(x)$ and $g(x)$ are continuous on the interval $[a, b]$, the functions $f(x)$, $g(x)$, $f^{2}(x)$, and $g^{2}(x)$ are integrable on $[a, b]$. Dividing the interval $[a, b]$ into $n$ equal parts and taking the left endpoint of each subinterval as $\xi_{i}$, by the definition of the definite ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,030 |
Proposition 4 If the series $\sum_{i=1}^{\infty}\left|a_{1 i}\right|^{m}, \sum_{i=1}^{\infty}\left|a_{2 i}\right|^{m}, \cdots \sum_{i=1}^{\infty}\left|a_{m i}\right|^{m}$ all converge, and for $\forall n \in N$ there is the inequality
$$\left(\sum_{i=1}^{n}\left|a_{1 i} a_{2 i} \cdots a_{m i}\right|\right)^{m} \leq \su... | Proof: Given functions $f_{j}(x)$ defined on the interval $[a, b]$ and continuous $(j \in N)$, divide the interval $[a, b]$ into $m$ equal parts, then the length of each subinterval is $\Delta x$. Take the left endpoint $\xi_{i} (i=1,2, \cdots, m)$ of each subinterval, then we have
$$\begin{array}{l}
\int_{a}^{b}\left|... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,031 |
Inference 1 Given $a_{i}(i=1,2, \cdots, n)$ are positive numbers, $x_{i} \in$ $\mathbf{R}(i=1,2, \cdots n)$ and $\sum_{i=1}^{*} a_{i}=1$, then $\sum_{i=1}^{n} a_{i} x_{i}^{2} \geqslant\left(\sum_{i=1}^{n} a_{i} x_{i}\right)^{2}$. | $$\begin{array}{l}
\text { Prove } \because a_{i} \in \mathbf{R}^{+}(i=1,2, \cdots n) \text {, and } \sum_{i=1}^{n} a_{i}=1 \text {, } \\
\therefore \sum_{i=1}^{n} a_{i} x_{i}^{2}=\left(\sum_{i=1}^{n} a_{i}\right) \cdot\left(\sum_{i=1}^{n} a_{i} x_{i}^{2}\right) \\
=\left[\sum_{i=1}^{n}\left(\sqrt{a_{i}}\right)^{2}\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,032 |
Example 4 Given real numbers $x, y$ satisfy $\left\{\begin{array}{l}x-y+2 \geqslant 0 \\ x+y-4 \geqslant 0 \\ 2 x-y-5 \leqslant 0\end{array}\right.$, find the maximum value of $z=|x+2 y-4|$. | Solve: First, draw the feasible region that satisfies the conditions, as shown in the shaded area in Figure 4. Convert the objective function $z=|x+2 y-4|$ to $z=\sqrt{5} - \frac{|x+2 y-4|}{\sqrt{1^{2}+2^{2}}}$, the problem is then reduced to finding the maximum value of $\sqrt{5}$ times the distance from points $(x, y... | 21 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,033 |
Inference 2 Given $a_{i}(i=1,2, \cdots n)$ are positive numbers, $x_{i} \in$ $\mathbf{R}(i=1,2, \cdots n)$, and $\sum_{i=1}^{n} a_{i}=1$,
then $\sum_{i=1}^{n} \frac{x_{i}^{2}}{a_{i}} \geqslant\left(\sum_{i=1}^{n} x_{i}\right)^{2}$. | Given $\because a_{i} \in \mathrm{R}^{+}(i=1,2, \cdots n)$, and $\sum_{i=1}^{n} a_{i}=1$,
$$\begin{aligned}
\therefore \sum_{i=1}^{n} \frac{x_{i}^{2}}{a_{i}} & =\left(\sum_{i=1}^{n} a_{i}\right) \cdot\left(\sum_{i=1}^{n} \frac{x_{i}^{2}}{a_{i}}\right) \\
& =\left[\sum_{i=1}^{n}\left(\sqrt{a_{i}}\right)^{2}\right] \cdot... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,034 |
For Example 1, prove: $\frac{a^{2}}{b+c-a}+\frac{b^{2}}{c+a-b}+\frac{c^{2}}{a+b-c}$ $\geqslant a+b+c$, where $a, b, c$ are the sides of $\triangle A B C$. | $$\geqslant \frac{1}{a+b+c} \cdot(a+b+c)^{2}=a+b+c$$
When and only when $\frac{\frac{x}{a+b+c}}{a}=\frac{\frac{y}{a+b+c}}{b}=\frac{\frac{z}{a+b+c}}{c}$, i.e., $a=b=c$, the equality holds.
Proof: Let $x=b+c-a, y=c+a-b, z=a+b-c$,
Since $a, b, c$ are the three sides of $\triangle ABC$,
$$\therefore x>0, y>0, z>0 \text{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,035 |
Example 1 (Chongqing 2008) The maximum value of the function $y=\sqrt{1-x}+$ $\sqrt{x+3}$ is $(\quad)$ | $$\begin{array}{l}
\text { Analysis: By Cauchy-Schwarz inequality, we have: } \\
y=\sqrt{1} \cdot \sqrt{1-x}+\sqrt{1} \cdot \sqrt{x+3} \leqslant \sqrt{1^{2}+1^{2}} . \\
\sqrt{(1-x)+(x+3)}=2 \sqrt{2} .
\end{array}$$
Therefore, the answer should be $2 \sqrt{2}$.
Evaluation: This problem involves an irrational function. ... | 2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,037 |
Example 2 (Zhejiang, 2008) If $\cos \alpha+2 \sin \alpha=-\sqrt{5}$, then $\tan \alpha=(\quad)$
A. $\frac{1}{2}$
B. 2
C. $-\frac{1}{2}$
D. -2 | By the Cauchy-Schwarz inequality, we have:
$$(\cos \alpha+2 \sin \alpha)^{2} \leqslant\left(1^{2}+2^{2}\right)\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)=$$
Given that $(\cos \alpha+2 \sin \alpha)^{2}=5$, we have $5 \leqslant 5$, which means the equality in the above inequality (1) holds. Therefore, by the conditio... | B | Algebra | MCQ | Yes | Yes | inequalities | false | 737,038 |
Example 3 (Hainan 2008) A certain edge of a geometric body has a length of $\sqrt{7}$. In the front view of the geometric body, the projection of this edge is a line segment with a length of $\sqrt{6}$. In the side view and top view of the geometric body, the projections of this edge are line segments with lengths of $... | Let the edge length of the geometric body be $\sqrt{7}$ for the edge $A C_{1}$. Construct a rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$ with $A C_{1}$ as the diagonal. Then the line segments $D C_{1}, B C_{1}$, and $A C$ are the projections of the edge $A C_{1}$ in the front view, side view, and top view of the... | C | Geometry | MCQ | Yes | Yes | inequalities | false | 737,039 |
Example 4 (2008 National I) If the line $\frac{x}{a}+\frac{y}{b}=1$ passes through the point $M(\cos \alpha, \sin \alpha)$, then ( )
A. $a^{2}+b^{2} \leqslant 1$
B. $a^{2}+b^{2} \geqslant 1$
C. $\frac{1}{a^{2}}+\frac{1}{b^{2}} \leqslant 1$
D. $\frac{1}{a^{2}}+\frac{1}{b^{2}} \geqslant 1$ | According to the conditions and Cauchy's inequality, we have
$$1=\left(\frac{\cos \alpha}{a}+\frac{\sin \alpha}{b}\right)^{2} \leqslant\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right) \cdot\left(\frac{1}{a^{2}}+\right.$$
$\left.\frac{1}{b^{2}}\right)$,
so $\frac{1}{a^{2}}+\frac{1}{b^{2}} \geqslant 1$. Therefore, the answ... | D | Geometry | MCQ | Yes | Yes | inequalities | false | 737,040 |
Example 5 (Shaanxi 2008) Given the sequence $\left\{a_{n}\right\}$ with the first term $a_{1} = \frac{3}{5}, a_{n+1} = \frac{3 a_{n}}{2 a_{n} + 1}, n=1,2, \cdots$. (1) Find the general term formula for $\left\{a_{n}\right\}$; (2) Prove that $a_{1} + a_{2} + \cdots + a_{n} > \frac{n^{2}}{n+1}$. | From the condition $a_{n+1}=\frac{3 a_{n}}{2 a_{n}+1} \Rightarrow \frac{1}{a_{n+1}}=\frac{1}{3} \cdot \frac{1}{a_{n}}+ \frac{2}{3}$, it is easy to find that the general term formula for $\left\{a_{n}\right\}$ is $a_{n}=\frac{3^{n}}{3^{n}+2}>0$, thus $\frac{1}{a_{n}}=1+\frac{2}{3^{n}}$.
By the Cauchy-Schwarz inequality... | a_{1}+a_{2}+\cdots+a_{n} > \frac{n^{2}}{n+1} | Algebra | proof | Yes | Yes | inequalities | false | 737,041 |
Example 1 Let $a, b>0$, if $\sqrt{a}+1>\sqrt{b}, x>1$, prove that $a x+$ $\frac{x}{x-1}>b$.
| Prove that the left side $=\left(a x+\frac{x}{x-1}\right)\left(\frac{1}{x}+\frac{x-1}{x}\right) \geqslant(\sqrt{a}+1)^{2}>b=$ right side. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,042 |
Example 2 Let $x, y>0, x+y=1$, prove that $\frac{1}{x}+\frac{2}{y} \geqslant 3+2 \sqrt{2}$. | $\begin{array}{l}\text { Prove } \frac{1}{x}+\frac{2}{y}=\left(\frac{1}{x}+\frac{2}{y}\right)(x+y) \geqslant(1+\sqrt{2})^{2}= \\ 3+2 \sqrt{2} \text {, that is, } \frac{1}{x}+\frac{2}{y} \geqslant 3+2 \sqrt{2} .\end{array}$ | \frac{1}{x}+\frac{2}{y} \geqslant 3+2 \sqrt{2} | Inequalities | proof | Yes | Yes | inequalities | false | 737,043 |
Example 3 Let $a, b \in \mathbf{R}, a^{2}+b^{2}=5$, prove that $-5 \leqslant a+2 b \leqslant 5$. | Prove $\because 5 \cdot 5=\left(1^{2}+2^{2}\right)\left(a^{2}+b^{2}\right) \geqslant(a+2 b)^{2}$, $\therefore-5 \leqslant a+2 b \leqslant 5$. | -5 \leqslant a+2 b \leqslant 5 | Inequalities | proof | Yes | Yes | inequalities | false | 737,044 |
Example 4 Given that $a, b$ are positive numbers, $x, y \in \mathbf{R}$, and $a+b=1$, prove: $a x^{2}+b y^{2} \geqslant(a x+b y)^{2}$.
---
The translation maintains the original text's line breaks and format. | Prove $a x^{2}+b y^{2}=\left(a x^{2}+b y^{2}\right)(a+b) \geqslant(a x+b y)^{2}$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,045 |
Example 5 Let $a, b, c>0, a+b+c=1$, prove that $\frac{1}{a}+\frac{1}{b}+$
$$\frac{1}{c} \geqslant 9 \text {. }$$ | Prove that the left side $=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(a+b+c) \geqslant(1+1+1)^{2}=9$, so the original inequality holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,046 |
Example 6 Let $x, y, z \in(0,1)$, prove that $\frac{1}{1-x+y}+\frac{1}{1-y+z}$
$$+\frac{1}{1-z+x} \geqslant 3$$ | Proof: Let the denominators on the left side be $a, b, c$, then $a, b, c > 0, a + b + c = 3$.
$\therefore$ Left side $=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{3}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\right.$ $\left.\frac{1}{c}\right) \geqslant \frac{1}{3}(1+1+1)^{2}=3$, hence the original inequality holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,047 |
Example 7 Let $-1<x, y<1$, prove that $\frac{1}{1-y^{2}}+\frac{1}{1-x^{2}} \geqslant$
$$\frac{2}{1-x y}$$ | Prove $\because-1<x, y<1, \therefore 1+x^{2}+x^{4}+\cdots=$
$$\begin{array}{l}
\frac{1}{1-x^{2}}, 1+y^{2}+y^{4}+\cdots=\frac{1}{1-y^{2}} . \quad \therefore \frac{1}{1-x^{2}}+\frac{1}{1-y^{2}}=2 \\
+\left(x^{2}+y^{2}\right)+\left(x^{4}+y^{4}\right)+\cdots \geqslant 2\left[1+x y+(x y)^{2}+\cdots\right] \\
=\frac{2}{1-x y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,048 |
Question 1 Given $f(x)=\sqrt{4+x^{2}}$, for any real numbers $a, b, a \neq b$, prove: $|f(a)-f(b)|<|a-b|$. | Let $a=2 \tan \alpha, b=2 \tan \beta, \alpha, \beta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, then we have $f(a)=\sqrt{4+4 \tan ^{2} \alpha}=\frac{2}{\cos \alpha}$, i.e., $f(a)=\frac{2}{\cos \alpha}$, similarly $f(b)=\frac{2}{\cos \beta}$.
Thus, the inequality to be proven is equivalent to
$\left|\frac{1}{\cos ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,049 |
Question 3 Given $a, b, c \in \mathbf{R},|a|<1,|b|<$ $1,|c|<1$, prove: $\left|\frac{a+b+c+a b c}{1+a b+b c+c a}\right|<1$.
| Using a binary conclusion to prove the ternary case, in fact,
from $|a|<1,|b|<1$, we get $\left|\frac{a+b}{1+a b}\right|<1$; from $\left|\frac{a+b}{1+a b}\right|<1, \quad | \quad c \quad |<1$, we get $\left|\frac{\frac{a+b}{1+a b}+c}{1+\frac{a+b}{1+a b} \cdot c}\right|<1$,
simplifying, we obtain the inequality to be p... | proof | Algebra | proof | Yes | Yes | inequalities | false | 737,050 |
Example 3 Solve the equation $3^{x}+4^{x}=5^{x}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
However, since the provided text is already in English, here is the confirmation of the text with the requested format preserved... | The original equation can be transformed into $\left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}=1$. Let $f(x)=\left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}$,
$\because f(x)$ is a monotonically decreasing function on $\mathbf{R}$, and $\left(\frac{3}{5}\right)^{2}+\left(\frac{4}{5}\right)^{2}=1$, i.... | x=2 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,051 |
Example 4 Solve the inequality $5 \sin ^{5} x+3 \sin ^{3} x>5 \cos ^{5} x+3 \cos ^{3} x$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Let $f(x)=5 x^{5}+3 x^{3}$, then the original inequality is $f(\sin x)>f(\cos x)$. Since $f(x)$ is a monotonically increasing function on $\mathbf{R}$, we have $\sin x>\cos x$.
Solving it, we get $2 k \pi+\frac{\pi}{4}<x<2 k \pi+\frac{5 \pi}{4}, k \in \mathbf{Z}$. | 2 k \pi+\frac{\pi}{4}<x<2 k \pi+\frac{5 \pi}{4}, k \in \mathbf{Z} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,052 |
Example 5 (1987 National College Entrance Examination Question) Suppose for all real numbers $x$, the inequality $x^{2} \cdot \log _{2} \frac{4(a+1)}{a}+2 x \log _{2} \frac{2 a}{a+1}+\log _{2} \frac{(a+1)^{2}}{4 a^{2}}$ $>0$ always holds. Find the range of values for $a$; | Let $\log _{2} \frac{(a+1)}{a}=t$,
then $x^{2} \cdot \log _{2} \frac{4(a+1)}{a}+2 x \log _{2} \frac{2 a}{a+1}+\log _{2} \frac{(a+1)^{2}}{4 a^{2}}>0$ always holds
$\Leftrightarrow(3+t) x^{2}-2 t x+2 t>0$ always holds
$\Leftrightarrow t>-\frac{3 x^{2}}{x^{2}-2 x+2}$
$\Leftrightarrow$ Let $f(x)=-\frac{3 x^{2}}{x^{2}-2 x+2... | a \in(0,1) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,053 |
Example 1 Let $x, y, z \in \mathbf{R}^{+}$, prove:
$$\sqrt{\frac{x}{y+z}}+\sqrt{\frac{y}{z+x}}+\sqrt{\frac{z}{x+y}} \geq 2 .$$ | Prove $\because \sqrt{\frac{x}{y+z}}=\frac{x}{\sqrt{x} \cdot \sqrt{y+z}}$
$$\geq \frac{x}{\frac{x+y+z}{2}}=\frac{2 x}{x+y+z} \cdots \cdots(1)$$
Similarly, we have: $\sqrt{\frac{y}{z+x}} \geq \frac{2 y}{x+y+z} \cdots \cdots(2)$
$$\sqrt{\frac{z}{x+y}} \geq \frac{2 z}{x+y+z} \cdots \cdots(3)$$
Adding the above three ine... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,054 |
Example 2 (Source unknown, excerpted from the Math Station Forum bbs.maths $168 . c o m$, solved by the author)
If $a, b, c>0$ and $a+b+c=1$, prove that:
$$\frac{3+a}{1+a^{2}}+\frac{3+b}{1+b^{2}}+\frac{3+c}{1+c^{2}} \leq 9$$ | $$\begin{array}{l}
\text { Prove } \because a, b, c>0 \text { and } a+b+c=1, \\
\therefore \frac{3+a}{1+a^{2}}+\frac{3}{10}(3 a-1)-3=\frac{(a-3)(3 a-1)^{2}}{10\left(1+a^{2}\right)} \leq 0, \\
\therefore \frac{3+a}{1+a^{2}}+\frac{3}{10}(3 a-1) \leq 3 \cdots(1) \\
\frac{3+b}{1+b^{2}}+\frac{3}{10}(3 b-1) \leq 3 \cdots(2) ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,055 |
Example 3 (39th IMO Shortlist) Let $x, y, z \in \mathbf{R}$ and $xyz=1$, prove that:
$$\frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)} \geq \frac{3}{4}.$$ | Analyzing the condition for equality, that is, when $x=y=z=1$,
we have $\frac{x^{3}}{(1+y)(1+z)}=\frac{1}{4}=\frac{1+y}{8}=\frac{1+z}{8}$,
so we can construct a local inequality.
Prove $\frac{x^{3}}{(1+y)(1+z)}+\frac{1+y}{8}+\frac{1+z}{8}$
$$\geq 3 \sqrt[3]{\frac{x^{3}}{(1+y)(1+z)} \cdot \frac{1+y}{8} \cdot \frac{1+z}... | \frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)} \geq \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 737,056 |
Variation 4 Given $a, b, c, d \in \mathbf{R}^{+}$, determine the monotonicity of the function $f(x)=a \sqrt{c-x}+b \sqrt{x+d}$. | Solve: From $c-x \geqslant 0, x+d \geqslant 0$, we get the domain of the original function as $-d \leqslant x \leqslant c$.
Differentiating the function $f(x)$, we get
$$f^{\prime}(x)=-\frac{a}{2 \sqrt{c-x}}+\frac{b}{2 \sqrt{x+d}}$$
From $f^{\prime}(x)=0$, we get $x=\frac{b^{2} c-a^{2} d}{a^{2}+b^{2}}$.
When $x \in\l... | f(x)=a \sqrt{c-x}+b \sqrt{x+d} \text{ is increasing on } \left[-d, \frac{b^{2} c-a^{2} d}{a^{2}+b^{2}}\right] \text{ and decreasing on } \left[\frac{b^{2} c-a^{2} d}{a^{2}+b^{2}}, c\right] | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,059 |
$$\begin{array}{l}
(1.1) I\left(m+x_{1}\right) \leqslant\left[m+\left(\sum \hat{x_{1}}\right) / n\right]^{n}\left(x_{i} \geqslant 0 \quad m>0\right) \\
(1.2) \Pi\left(m+x_{1}\right)^{p} \leqslant m+q_{q_{1}}\left(m>0 \quad x_{1} \geqslant 0 \quad q_{i} \geqslant 0 \quad \Sigma q_{1}=1\right) \\
(1.3) \Pi\left(m+x_{1}\r... | To prove that since (1.2) is a special case of $(1.3)$ when $\Sigma_{\mathrm{p}}=1$, and (1.1) is a special case of (1.2) when $q=\frac{1}{n}$, we only need to prove $(1.3)$. Construct the auxiliary function $f(x)=\ln (m+x)(m>0)$, and we have $f^{\prime}(x)=-1 /(m+x)^{2}-m$ holds. By Jensen's inequality $(2)^{\prime}$,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,060 |
Example 1 Determine the positional relationship between a line and an ellipse or hyperbola.
(1) $l: 2 x-y-8=0, c: \frac{x^{2}}{2}+y^{2}=1$;
(2) $l: x+y-2=0, c: \frac{x^{2}}{4}-y^{2}=1$. | Solution: (1) It is easy to know that the two foci of the ellipse are $F_{1}(-1,0)$, $F_{2}(1,0)$, and the semi-minor axis $b=1$.
$$\because[2 \cdot(-1)-0-8] \cdot(2 \times 1-0-8)>0 \text {, }$$
$\therefore F_{1} 、 F_{2}$ are on the same side of $l$.
Also, $d_{1} \cdot d_{2}=\frac{|-2-8|}{\sqrt{5}} \cdot \frac{|2-8|}{\... | not found | Geometry | math-word-problem | Yes | Yes | inequalities | false | 737,065 |
Example 2 Suppose the foci of an ellipse are the same as those of the hyperbola $4 x^{2}-5 y^{2}=$ 20, and it is tangent to the line $x-y+9=0$. Find the standard equation of this ellipse. | Solution: It is easy to know that the two foci of the ellipse are $F_{1}(-3,0)$, $F_{2}(3,0)$, and the lengths of the semi-major axis and semi-minor axis are $a$ and $b$ respectively,
$\because$ the line is tangent to the ellipse,
$\therefore d_{1} \cdot d_{2}=\frac{|-3+9|}{\sqrt{2}} \cdot \frac{|3+9|}{\sqrt{2}}=36=b^{... | \frac{x^{2}}{45}+\frac{y^{2}}{36}=1 | Geometry | math-word-problem | Yes | Yes | inequalities | false | 737,066 |
Example 7 Prove: $\sin ^{10} \alpha+\cos ^{10} \alpha \geqslant \frac{1}{16}$. | Proof: $\because \sin ^{2} \alpha+\cos ^{2} \alpha=1$, and the function $f(x)=$ $x^{5}$ is convex on $[0,1]$, hence by Jensen's inequality we have
$$\begin{aligned}
{\left[\sin ^{2} \alpha\right]^{5}+\left[\cos ^{2} \alpha\right]^{5} } & \geqslant 2 \cdot\left[\frac{\sin ^{2} \alpha+\cos ^{2} \alpha}{2}\right]^{5} \\
&... | \frac{1}{16} | Inequalities | proof | Yes | Yes | inequalities | false | 737,068 |
Example 8 In $\triangle A B C$, prove that: $h_{a}+h_{b}+h_{c} \geqslant$ $9 r$. Where $h_{a}, h_{b}, h_{c}$ are the three altitudes of the triangle, and $r$ is the radius of the inscribed circle. | Proof: Let the area of this triangle be $S_{\Delta}$, then
$$h_{a}+h_{b}+h_{c}=\frac{2 S_{\Delta}}{a}+\frac{2 S_{\Delta}}{b}+\frac{2 S_{\Delta}}{c}=\frac{1}{\frac{a}{2 S_{\Delta}}}$$
$+\frac{1}{\frac{b}{2 S_{\Delta}}}+\frac{1}{\frac{c}{2 S_{\Delta}}}$, clearly the function $f(x)=\frac{1}{x}$ is convex on $(0, +\infty)$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,069 |
Example 9 Let $m_{a}, m_{b}, m_{c}$ be the three medians of $\triangle A B C$, prove that: $m_{a}^{2}+m_{b}^{2}+m_{c}^{2} \geqslant s^{2}$ (where $s$ is the semiperimeter of $\triangle A B C$). | Proof: From Apollonius's theorem: $4 m_{a}^{2}=2 b^{2}+2 c^{2}$
$$\begin{array}{l}
-a^{2}, \text { we get } 4\left(m_{a}^{2}+m_{b}^{2}+m_{c}^{2}\right)=\left(2 b^{2}+2 c^{2}-a^{2}\right) \\
+\left(2 c^{2}+2 a^{2}-b^{2}\right)+\left(2 a^{2}+2 b^{2}-c^{2}\right)=3 a^{2}+ \\
3 b^{2}+3 c^{2} .
\end{array}$$
Obviously, the... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,070 |
Example 10 In $\triangle A B C$, prove: $a^{4}+b^{4}+c^{4} \geqslant$ $16 S_{\Delta}^{2}$. | Proof: By Weitzenböck's inequality, we know:
\[a^{2}+b^{2}+c^{2} \geqslant 4 \sqrt{3} S_{\Delta},\]
The function \(f(x)=x^{2}\) is convex on \((0,+\infty)\), hence we have
\[
\begin{array}{l}
a^{4}+b^{4}+c^{4} \geqslant 3 \cdot\left(\frac{a^{2}+b^{2}+c^{2}}{3}\right)^{2} \\
=3 \cdot\left(\frac{4 \sqrt{3} S_{\Delta}}{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,071 |
Example 1 The table below gives an "arithmetic array":
where each row and each column are arithmetic sequences, $a_{i j}$ represents the number located at the $i$-th row and $j$-th column.
(I) Write down the value of $a_{45}$;
(II) Write down the formula for $a_{i i}$;
(III) Prove: A positive integer $N$ is in this ari... | Solution: (I) $a_{45}=49$.
(II) Grouping by rows, the first row, or the first group, is an arithmetic sequence with the first term 4 and a common difference of 3: $a_{ij} = 4 + 3(j-1)$,
the second group is an arithmetic sequence with the first term 7 and a common difference of 5:
$$a_{2j} = 7 + 5(j-1),$$
$\cdots$,
the... | a_{ij} = i(2j + 1) + j | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,072 |
Example 3 Given that the center of the hyperbola is at the origin, the foci are on the $x$-axis, and the hyperbola is tangent to the lines $2 x-y+\sqrt{21}=0$ and $x+y+\sqrt{3}=0$. Find the equation of this hyperbola. | Solution: Let the two foci of the hyperbola be $F_{1}(-c, 0)$, $F_{2}(c, 0)$, and the imaginary semi-axis length be $b$,
$\because$ The line $2 x-y+\sqrt{21}=0$ is tangent to the hyperbola,
$$\therefore\left\{\begin{array}{l}
(-2 c+\sqrt{21})(2 c+\sqrt{21})<0 \\
\frac{|-2 c+\sqrt{21}|}{\sqrt{5}} \cdot \frac{|2 c+\sqrt{... | \frac{x^{2}}{6}-\frac{y^{2}}{3}=1 | Geometry | math-word-problem | Yes | Yes | inequalities | false | 737,073 |
Inference: Let $x_{i} \in[a, b](i=1,2,3, \cdots n)$ and not all equal, for a positive function $f(x)$ defined on $[a, b]$,
$1^{\circ}$ If $f(x)$ is a convex function on $[a, b]$, then $f^{n}\left(\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) \geqslant f\left(x_{1}\right) f\left(x_{2}\right) \cdots f\left(x_{n}\right)$;
$2^... | Prove: Since $f(x)>0$ and is a convex function on $[a, b]$, for any two different points $x_{1}, x_{2}$ on $[a, b]$
there always holds $f\left(\frac{x_{1}+x_{2}}{2}\right) \geqslant \frac{f\left(x_{1}\right)+f\left(x_{2}\right)}{2}$.
$$\begin{array}{l}
\quad \text { Let } F(x)=\lg f(x), \\
\quad \because F\left(\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,075 |
Example 1 If $a_{1}+a_{2}+\cdots+a_{n}=1\left(a_{i}>0\right)$, prove: $\left(1+\frac{1}{a_{1}}\right)\left(1+\frac{1}{a_{2}}\right) \cdots\left(1+\frac{1}{a_{n}}\right)>(1+n)^{n}$. | Proof: Since $f(x)=1+\frac{1}{x}$ is a convex function on $(0,+\infty)$, by the corollary, for any $a_{i} \in(0,+\infty)$ $(i=1,2, \cdots, n)$,
we always have: $\left(1+\frac{1}{a_{1}}\right)\left(1+\frac{1}{a_{2}}\right) \cdots\left(1+\frac{1}{a_{n}}\right)>(1+$ $\left.\frac{1}{\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}}\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,076 |
Example 2 Let $a^{2}+b^{2}+c^{2}=8\left(a, b, c \in R^{+}\right)$, prove: $a^{3}+b^{3}+c^{3} \geqslant \frac{16 \sqrt{6}}{3}$. | Prove: The function $f(x)=x^{\frac{3}{2}}$ is obviously convex on $(0,+\infty)$. Therefore, by Jensen's inequality, we have:
$$\begin{aligned}
& \left(a^{2}\right)^{\frac{3}{2}}+\left(b^{2}\right)^{\frac{3}{2}}+\left(c^{2}\right)^{\frac{3}{2}} \geqslant 3\left(\frac{a^{2}+b^{2}+c^{2}}{3}\right)^{\frac{3}{2}} \\
= & 3\l... | a^{3}+b^{3}+c^{3} \geqslant \frac{16 \sqrt{6}}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 737,077 |
Example 3 If $a+b+c=1$, prove that: $\sqrt{13 a+1}+$
$$\sqrt{13 b+1}+\sqrt{13 c+1} \leqslant 4 \sqrt{3} .$$ | Proof: Clearly, the function $f(x)=\sqrt{13 x+1}$ is convex on $\left(-\frac{1}{13},+\infty\right)$, so by Jensen's inequality we have
$$\begin{array}{l}
\quad \sqrt{13 a+1}+\sqrt{13 b+1}+\sqrt{13 c+1} \leqslant 3 \cdot \\
\sqrt{13 \cdot \frac{a+b+c}{3}+1}=4 \sqrt{3} .
\end{array}$$ | 4 \sqrt{3} | Inequalities | proof | Yes | Yes | inequalities | false | 737,078 |
Example 4 Given $x+2 y+4 z=12$, prove:
$$\sqrt{3 x+4}+\sqrt{6 y+4}+\sqrt{12 z+4} \leqslant 12 .$$ | Proof: Clearly, the function $f(t)=\sqrt{t}$ is convex on $(0,+\infty)$, so by Jensen's inequality we have
$$\begin{array}{l}
\quad \sqrt{3 x+4}+\sqrt{6 y+4}+\sqrt{12 z+4} \leqslant 3 \cdot \\
\sqrt{\frac{(3 x+4)+(6 y+4)+(12 z+4)}{3}} \\
\quad=3 \cdot \sqrt{\frac{3(x+2 y+4 z)+12}{3}} \\
\quad=12 .
\end{array}$$ | 12 | Inequalities | proof | Yes | Yes | inequalities | false | 737,079 |
Example 5 In $\triangle A B C$, prove that: $\sin A+\sin B+\sin C \leqslant \frac{3 \sqrt{3}}{2}$ | Proof: Clearly, the function $f(x)=\sin x$ is concave in $\left(0, \frac{\pi}{2}\right)$, so by Jensen's inequality we have
$$\begin{array}{l}
\sin A+\sin B+\sin C \leqslant 3 \sin \frac{A+B+C}{3}= \\
3 \sin \frac{\pi}{3}=\frac{3 \sqrt{3}}{2}
\end{array}$$ | \frac{3 \sqrt{3}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 737,080 |
Theorem (Jensen's Inequality): If the function $f(x)$ has a second derivative on the interval $I$, and for all $x \in I$, we have $f^{\prime \prime}(x) \geqslant 0$, then
$$f\left(\sum_{i=1}^{n} q_{1} x_{1}\right) \leqslant \sum_{i=1}^{n} q_{i} f\left(x_{i}\right)$$
where $x_{i} \in I, q_{i}>0, i=1,2, \cdots, n$, and ... | $$\begin{array}{c}
\text { Proof: Let } x_{0}=\sum_{i=1}^{n} q_{i} x_{i} \in I, \text { by Taylor's formula, we have } \\
f\left(x_{i}\right)=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x_{i}-x_{0}\right)+\frac{f^{\prime \prime}\left(\xi_{i}\right)}{2!} \\
\left(x_{i}-x_{0}\right)^{2}
\end{array}$$
where $\x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,081 |
Inference 2 If $x_{i}>0, q_{i}>0, i=1,2, \cdots, n$, and $\sum_{i=1}^{n} q_{i}=1, \forall p>1$, then we have
$$\sum_{i=1}^{n} q_{i} x_{i} \frac{1}{\rho} \leqslant\left(\sum_{i=1}^{n} q_{i} x_{i}\right)^{\frac{1}{p}}$$ | Prove that by Jensen's inequality (1), we have
$$f\left(\sum_{i=1}^{n} q_{i} x_{i}^{\frac{1}{p}}\right) \leqslant \sum_{i=1}^{n} q_{i} f\left(x_{i} \frac{1}{p}\right)$$
Taking \( f(x) = x^p, x > 0, f'(x) = p x^{p-1}, f''(x) = p(p-1) x^{p-2} \), since \( p > 1 \), for all \( x > 0 \), we have \( f''(x) > 0 \). By equat... | \sum_{i=1}^{n} q_{i} x_{i}^{\frac{1}{p}} \leqslant \left(\sum_{i=1}^{n} q_{i} x_{i}\right)^{\frac{1}{p}} | Inequalities | proof | Yes | Yes | inequalities | false | 737,083 |
Inference 3 If $x_{i}>0, q_{i}>0, i=1,2, \cdots, n$, and $\sum_{i=1}^{n} q_{i}=1$, then
$$\prod_{i=1}^{n} x_{i}^{q_{i}} \leqslant \sum_{i=1}^{n} q_{i} x_{i}$$ | Given $x_{i}>0, q_{i}>0$, let $y_{i}=\ln x_{i}$ or $x_{i}=e^{y_{i}}$, then $x_{i}^{q_{i}}=e^{q_{i} \ln x_{i}}=e^{q_{i} y_{i}}$. Taking $f(y)=e^{y} \forall y \in R$, we have $f^{\prime}(y)=f^{\prime \prime}(y)=e^{y}>0$. By Jensen's inequality (1), we have
$$\begin{aligned}
\prod_{i=1}^{n} x_{i}^{q_{i}} & =\prod_{i=1}^{n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,084 |
Example 1 Prove: When $a_{i}>0$, $i=1,2, \cdots, n$, the inequality
$$\begin{array}{l}
\frac{n}{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}} \leqslant \sqrt[n]{a_{1} a_{2} \cdots a_{n}} \leqslant \\
\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}
\end{array}$$
holds. Here, the equality holds if and only if $a_{1}=a_{2}=... | Proof: Let $f(x)=-\ln x, \forall x \in(0, \infty)$, then $f^{\prime \prime}(x)=\frac{1}{x^{2}}>0$, hence, the function $f(x)=-\ln x$ is strictly convex on $(0,+\infty)$. According to Corollary 1, taking $x_{i}=a_{i} \in(0,+\infty), q_{i}=\frac{1}{n}, i=1,2, \cdots, n, \sum_{i=1}^{n} q_{i}=1$, we have
$$-\ln \left(\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,085 |
Example 2 Let $g(s)=\left(\frac{1}{n} \sum_{i=1}^{n} x^{s}\right)^{\frac{1}{s}}$, for $\forall t, s \in R$, and $t<0<s$, we have
$$g(t) \leqslant g(0) \leqslant g(s)$$ | Proof Consider $n$ positive numbers: $x_{1}^{s}, x_{2}^{s}, \cdots, x_{n}^{s}$. From Example 1, we have
$$\sqrt[n]{\prod_{i=1}^{n} x_{i}^{s}} \leqslant \frac{1}{n} \sum_{i=1}^{n} x_{i}^{s}$$
or
$$\sqrt[n]{\prod_{i=1}^{n} x_{i}} \leqslant\left(\frac{1}{n} \sum_{i=1}^{n} x_{i}^{s}\right)^{\frac{1}{n}}$$
i.e. $\square$
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,086 |
Example 3 (Hölder's Inequality) If $a_{i}>0, b_{i}>0, i=$ $1,2, \cdots, n$, and $p>1, \frac{1}{p}+\frac{1}{q}=1$, then
$$\sum_{i=1}^{n} a_{i} b_{i} \leqslant\left(\sum_{i=1}^{n} a_{i}^{p}\right)^{\frac{1}{p}}\left(\sum_{i=1}^{n} b_{i}^{q}\right)^{\frac{1}{q}}$$ | Prove that for the function: $f(x)=-x^{\frac{1}{q}}, \forall x>0$, we have
$$f^{\prime \prime}(x)=-\frac{1}{q}\left(\frac{1}{q}-1\right) x^{\frac{1}{q}-2}>0$$
Let $x_{i}=\frac{b_{i}^{q}}{a_{i}^{p}}, q_{i}=\frac{a_{i}^{p}}{\sum_{i=1}^{n} a_{i}^{p}}$. Clearly, $\sum_{i=1}^{n} q_{i}=1$, and using $\frac{1}{p}+\frac{1}{q}... | \sum_{i=1}^{n} a_{i} b_{i} \leqslant\left(\sum_{i=1}^{n} a_{i}^{p}\right)^{\frac{1}{p}}\left(\sum_{i=1}^{n} b_{i}^{q}\right)^{\frac{1}{q}} | Inequalities | proof | Yes | Yes | inequalities | false | 737,087 |
Example 4 (Minkowski Inequality) If $a_{i}>0, b_{i}>0, i$ $=1,2, \cdots, n$, and $p>1, \frac{1}{p}+\frac{1}{q}=1$, then we have $\left(\sum_{i=1}^{n}\left(a_{i}+b_{i}\right)^{p}\right)^{\frac{1}{p}} \leqslant\left(\sum_{i=1}^{n}\left(a_{i}^{p}\right)^{\frac{1}{p}}+\left(\sum_{i=1}^{n} b_{i}^{p}\right)^{\frac{1}{p}}\rig... | $$\begin{array}{l}
\sum_{i=1}^{n}\left(a_{i}+b_{i}\right)^{p}=\sum_{i=1}^{n} a_{i}\left(a_{i}+b_{i}\right)^{p-1}+\sum_{i=1}^{n} b_{i}\left(a_{i}+ \right. \\
\left.b_{i}\right)^{p-1}
\end{array}$$
Applying Hölder's inequality to the two sums on the right-hand side, we get
$$\begin{array}{l}
\quad \sum_{i=1}^{n}\left(a_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,088 |
Example I Let $x_{k}>0 \quad(k=1,2,3, \cdots, n)$, prove that:
$$\sqrt[n]{x_{1} x_{2} \cdots x_{n}} \leqslant \frac{x_{1}+x_{2}+\cdots+x_{n}}{n}$$ | Prove: Introduce $f(x)=-\ln x(x \in(0,+\infty))$,
$$f^{\prime \prime}(x)=\frac{1}{x^{2}}>0 \quad(x \in(0,+\infty))$$
According to Jensen's inequality:
$$\begin{array}{l}
-\ln \frac{x_{1}+x_{2}+\cdots+x_{n}}{n} \leqslant-\frac{1}{n}\left(\ln x_{1}+\right. \\
\left.\ln x_{2}+\cdots+\ln x_{n}\right) \\
\ln \sqrt[n]{x_{1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,090 |
Example 2 Prove: $\left(a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}\right)^{2}$
$$\begin{array}{l}
\leqslant\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+\cdots+\right. \\
\left.b_{n}^{2}\right)
\end{array}$$ | Prove: Introduce $f(x)=x^{2} \quad(x \in(-\infty, +\infty))$ then:
$$f^{\prime \prime}(x)=2>0, (x \in(-\infty,+\infty))$$
According to Jensen's inequality:
$$\begin{array}{l}
\left(p_{1} x_{1}+p_{2} x_{2}+\cdots+p_{n} x_{n}\right)^{2} \\
\leqslant p_{1} x_{1}^{2}+p_{2} x_{2}^{2}+\cdots+p_{n} x_{n}^{2}
\end{array}$$
E... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,093 |
Example 3 In $\triangle A B C$, prove:
$$\sin A+\sin B+\sin C \leqslant \frac{3}{2} \sqrt{3}$$ | Prove: Let $f(x)=-\sin x \quad(0 \leqslant x<\pi)$, according to Jensen's inequality:
$-\sin \frac{A+B+C}{3} \leqslant \frac{1}{3}(-\sin A-\sin B-\sin C)$
i.e., $\sin A+\sin B+\sin C \leqslant \frac{3}{2} \sqrt{3}$.
Equality holds if and only if $A=B=C=\frac{\pi}{3}$. | \sin A+\sin B+\sin C \leqslant \frac{3}{2} \sqrt{3} | Inequalities | proof | Yes | Yes | inequalities | false | 737,094 |
Example 4 Let $a_{k}>0,(k=1,2, \cdots, n)$, and $\alpha$ $>1$.
Prove: $\left(\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)^{a} \leqslant \frac{1}{n}\left(a_{1}{ }^{a}+\right.$ $\left.a_{2}{ }^{\alpha}+\cdots+a_{n}{ }^{a}\right)$ | Prove: Introduce $f(x)=x^{a}(x>0)$, then
$$f^{\prime \prime}(x)=\alpha(a-1) x^{\alpha-2}>0,(x>0)$$
According to Jensen's inequality
$$\begin{array}{l}
\left(\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)^{a} \leqslant \frac{1}{n}\left(a_{1}^{\alpha}+a_{2}^{a}+\cdots+\right. \\
\left.a_{n}^{a}\right) .
\end{array}$$
Equali... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,095 |
Example 5 Let $a_{k}>0,(k=1,2,3, \cdots, n)$, and $a_{1}+a_{2}+\cdots+a_{n}=A$
Prove: $\left(a_{1}+\frac{1}{a_{1}}\right)^{2}+\left(a_{2}+\frac{1}{a_{2}}\right)^{2}+\cdots+$ $\left(a_{n}+\frac{1}{a_{n}}\right)^{2} \geqslant n\left(\frac{A}{n}+\frac{n}{A}\right)^{2}$. | Prove: Introduce $f(x)=\left(x+\frac{1}{x}\right)^{2} \quad(x>0)$, then $f^{\prime \prime}(x)=2+\frac{6}{x^{4}}>0 \quad(x>0)$
According to Jensen's inequality:
$$\begin{array}{l}
\left(\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}+\frac{n}{a_{1}+a_{2}+\cdots+a_{n}}\right)^{2} \\
\leqslant \frac{1}{n}\left[\left(a_{1}+\frac{1}{a_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,096 |
Original: $f\left(\frac{2000}{2001}\right)+f\left(\frac{1999}{2001}\right)+\cdots+f\left(\frac{1}{2001}\right)$. | $\begin{array}{l}\text { Sol } \because f(x)+f(1-x)=\frac{a^{2 x}}{a^{2 x}+a}+ \\ \frac{a^{2(1-x)}}{a^{2(1-x)}+a}=\frac{a^{2 x}}{a^{2 x}+a}+\frac{a}{a^{2 x}+a}=1 \\ \quad \therefore\left[f\left(\frac{1}{2001}\right)+f\left(\frac{2000}{2001}\right)\right]+\left[f\left(\frac{2}{2001}\right)+\right. \\ \left.f\left(\frac{... | 1000 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,097 |
Example 9 Given $a, b \in R^{+}$, and $\frac{1}{a}+\frac{1}{b}=1$, prove that for every natural number $n$. $(a+b)^{n}-a^{n}-b^{n}$ $\geqslant 2^{2 n}-2^{n+1}$. (1988 National High School Mathematics League Question) | Prove: Let $A=(a+b)^{n}-a^{n}-b^{n}$
$$=\mathrm{C}_{n}^{1} a^{n-1} b+\mathrm{C}_{n}^{2} a^{n-2} b^{2}+\cdots$$
$+\mathrm{C}_{n}^{n-1} a b^{n-1}$, then we also have
$$A=\mathrm{C}_{n}^{n-1} a b^{n-1}+\mathrm{C}_{n}^{n-2} a^{2} b^{n-2}+\cdots+\mathrm{C}_{n}^{1} a^{n-1}$$
$b$. Adding the two equations, we get
$$\begin{ali... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,098 |
\[
\begin{array}{l}
\quad \text { Let } a_{k}>0, (k=1,2, \cdots, m, \text { and } m \geqslant 2, m \\
\in N, a_{1}+a_{2}+\cdots+a_{n}=A \text {, then }: \sum_{k=1}^{n}\left(a_{k}+ \\
\left.\frac{1}{a_{k}}\right)^{m} \geqslant n\left(\frac{A}{n}+\frac{n}{A}\right)^{m} . \quad \text { (*) }
\end{array}
\] | Prove: Introduce the function $f(x)=\left(x+\frac{1}{x}\right)^{m} \quad(x>0)$
Then $f^{\prime \prime}(x)=m\left(x+\frac{1}{x}\right)^{m-2}[(m-1)(1-$
$$\begin{array}{l}
\left.\left.\frac{1}{x^{2}}\right)^{2}+2\left(x+\frac{1}{x}\right) x^{-3}\right] \\
\because(m-1)\left(1-\frac{1}{x^{2}}\right)^{2}+2\left(x+\frac{1}{x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,099 |
Text [1] proposes to use the properties of convex functions and Jensen's inequality:
$$-\ln \left(\sum_{j=1}^{n} \frac{a_{j}}{\sum_{i=1}^{n} a_{i}} \cdot \frac{x_{j}}{a_{j}}\right) \leqslant-\sum_{j=1}^{n} \frac{a_{j}}{\sum_{i=1}^{n} a_{i}} \cdot \ln \frac{x_{j}}{a_{j}}$$ | Construct the following problem: Find the maximum value of the function $u=\sum_{i=1}^{n} a_{i} \ln y_{i}$ on the sphere $\sum_{i=1}^{n} y_{i}^{2}=m r^{2}$, where $a_{i} \in \mathbb{R}^{+}, y_{i} \in \mathbb{R}^{+}$, and $m=\sum_{i=1}^{n} a_{i}$. Construct the Lagrangian function
$$L=\sum_{i=1}^{n} a_{i} \ln y_{i}+\lam... | null | Inequalities | proof | Yes | Yes | inequalities | false | 737,100 |
Proposition 2 If $x_{i}, a_{i} \in R^{+} \quad(i=1,2, \cdots, n)$, then
$$\left(\prod_{i=1}^{n}\left(\frac{x_{i}}{a_{i}}\right)^{a_{i}}\right)^{\frac{1}{\sum_{i=1}^{n} a_{i}}} \leqslant \frac{\sum_{i=1}^{n} x_{i}}{\sum_{i=1}^{n} a_{i}} \leqslant\left(\prod_{i=1}^{n}\left(\frac{x_{i}}{a_{i}}\right)^{x_{i}}\right)^{\frac... | Prove that Proposition 1 easily leads to the establishment of the left inequality.
Since $\prod_{i=1}^{n} x_{i}^{a_{i}} \leqslant \prod_{i=1}^{n} a_{i}^{\sigma_{i}}\left(\frac{\sum_{i=1}^{n} x_{i}}{\sum_{i=1}^{n} a_{i}}\right)^{\sum_{i=1}^{n} a_{i}}$
$\Leftrightarrow \quad \prod_{i=1}^{n} a_{i}^{a_{i}} \geqslant \prod_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,106 |
The implicit function $z=z(x, y)$ (satisfying the conditions of the implicit function theorem), and considering the target function $f(x, y, z)=x y z(x, y)=F(x, y)$ as a composite function of $f$ and $z=z(x, y)$. This way, we can apply the sufficient conditions for extrema to make a judgment. For this purpose, the foll... | Let $G(x, y, z)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{a}$, then
thus
$$G_{x}=-\frac{1}{x^{2}}, G_{y}=-\frac{1}{y^{2}}, G_{z}=-\frac{1}{z^{2}}, z_{x}=-\frac{G_{x}}{G_{z}}=-\frac{z^{2}}{x^{2}}, z_{y}=-\frac{z^{2}}{y^{2}}$$
$$\begin{array}{c}
F_{x}=y z+x y z_{x}=y z-\frac{y z^{2}}{x}, F_{y}=x z+x y z_{y}=x z-\frac... | 27 a^{3} | Calculus | math-word-problem | Yes | Yes | inequalities | false | 737,107 |
Original problem (39th IMO Shortlist) Let $x, y, z$ be positive real numbers, and $x y z=1$, prove: $\quad \frac{x^{3}}{(1+y)(1+z)}+$ $\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)} \geqslant \frac{3}{4}$. | Proof: Let (1) the left side be $P$. By the AM-GM inequality, we easily get
$$\frac{8 x^{3}}{(1+y)(1+z)}+(1+y)+(1+z) \geqslant 6 x$$
By cyclically permuting $x, y, z$, we obtain the other two inequalities. Adding these three inequalities, we get $8 P+6+$ $2(x+y+z) \geqslant 6(x+y+z)$, hence $4 P \geqslant 2(x+y+z)$ $-... | P \geqslant \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 737,108 |
Promotion Proposition Let $n$ be a positive integer, $n \geqslant 2, x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers, and $x_{1} x_{2} \cdots x_{n}=1,\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots(1+ \left.x_{n}\right)=Q$, then $\quad \sum_{i=1}^{n} \frac{x_{i}^{n}\left(1+x_{i}\right)}{Q} \geqslant \frac{n}{2^{n... | Proof: Let (2) left be $P$. By the $n$-variable mean inequality, it is easy to get $2^{n} x_{i}^{n}\left(1+x_{i}\right) Q^{-1}+\left(1+x_{1}\right)+\cdots+\left(1+x_{i-1}\right)+$ $\left(1+x_{i+1}\right)+\cdots+\left(1+x_{n}\right) \geqslant 2 n x_{i}, i=1,2, \cdots, n$. Adding $n$ equations, we get $2^{n} P+n(n-1)+(n-... | P \geqslant n \cdot 2^{1-n} | Inequalities | proof | Yes | Yes | inequalities | false | 737,109 |
Example 6 Given a line segment $A B$ and a line $C D$ parallel to $A B$, let $n$ be a positive integer, $n \geqslant 2$. Using only a straightedge, construct a point that divides $A B$ into $n$ equal parts.
Construct a point that divides $A B$ into $n$ equal parts using only a straightedge, given a line segment $A B$ ... | Proof: By mathematical induction. When $n=2$, the proposition holds. Suppose point $I$ is the $(n-1)$-th equal division point of $AB$ (closest to point $B$) for $n \geqslant 3$. Connect $IG$ to intersect $BF$ at $J$, and extend $EJ$ to intersect $AB$ at $K$. Then,
\[
\frac{AK}{BK} = \frac{\triangle AEJ}{\triangle BEJ}... | proof | Geometry | math-word-problem | Yes | Yes | inequalities | false | 737,110 |
Example 1: Prove that in $\triangle A B C$, $\sin A+\sin B+\sin C \leqslant \frac{3 \sqrt{3}}{2}$. (The proof can be done using the convexity of $y=\sin x$ on $(0, \pi)$, which is omitted here.)
Similarly, I. In $\triangle A B C$, we have (1) $\cos A+\cos B+\cos C \leqslant \frac{3}{2}$; (2) $\cos \frac{A}{2} \cos \fr... | Prove: As shown in Figure 1, mark each angle, then \(\alpha + \beta + \gamma + \alpha' + \beta' + \gamma' = \pi\) and \(\alpha, \beta, \gamma, \alpha', \beta', \gamma' > 0\). In \(\triangle APB\), \(\triangle BPC\), \(\triangle CPA\), by the Law of Sines, we get \(PA \sin \alpha = PB \sin \beta'\), \(PB \sin \beta = PC... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,111 |
Example 2 (2005 National Paper I) Let positive numbers $p_{1}, p_{2}, \cdots, p_{2^{n}}$ satisfy $p_{1}+p_{2}+\cdots+p_{2^{n}}=1$. | Prove: $p_{1} \log _{2} p_{1}+p_{2} \log _{2} p_{2}+\cdots+p_{2^{n}} \geqslant-n$.
Solution: Let $g(x)=x \log _{2} x$, then $g^{\prime}(x)=\log _{2} x+\log _{2} e, g^{\prime \prime}(x)=\frac{1}{x} \log _{2} e>0$.
$\therefore g(x)$ is a convex function on $(0,+\infty)$, by Jensen's inequality we get $\sum_{k=1}^{2^{n}}... | proof | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,112 |
Example 3 (2011 Hubei Paper) Let $a_{k}, b_{k}(k=1,2, \cdots, n)$ be positive numbers. Prove:
(1) If $a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n} \leqslant b_{1}+b_{2}+\cdots +b_{n}$, then $a_{1}^{h} a_{2}^{h} \cdots a_{n}^{b} \leqslant 1$.
(2) If $b_{1}+b_{2}+\cdots+b_{n}=1$, then $\frac{1}{n} \leqslant b_{1}^{b_{1}^{b... | Proof (1) Let $\sum_{k=1}^{n} b_{k}=S$, then $\frac{\sum_{k=1}^{n} a_{k} b_{k}}{S} \leqslant$ 1. By the convexity of $y=\ln x$ on $(0,+\infty)$ and Jensen's inequality, we get $\sum_{k=1}^{n} \frac{b_{k}}{S} \ln a_{k} \leqslant \ln \sum_{k=1}^{n} \frac{b_{k} a_{k}}{S} \leqslant 0$. Therefore, $\prod_{k-1}^{n} a_{k}^{k}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,113 |
Example 4 (2003 National Competition) Let $\frac{3}{2} \leqslant r \leqslant 5$, prove:
$$2 \sqrt{x+1}+\sqrt{2 x-3}+\sqrt{15-3 x}<2 \sqrt{19} .$$ | Prove: Since $y=\sqrt{x}$ is a convex function on $(0,+\infty)$, then $2 \sqrt{x+1}+\sqrt{2 x-3}+\sqrt{15-3 x}=\sqrt{x+1}+$ $\sqrt{x+1}+\sqrt{2 x-3}+\sqrt{15-3 x} \leqslant 4$ $\sqrt{\frac{2(x+1)+(2 x-3)+(15-3 x)}{4}}=2 \sqrt{x+14}$ $\leqslant 2 \sqrt{19}$, and $x+1,2 x-3,15-3 x$ cannot be equal simultaneously, so the ... | 2 \sqrt{x+1}+\sqrt{2 x-3}+\sqrt{15-3 x}<2 \sqrt{19} | Inequalities | proof | Yes | Yes | inequalities | false | 737,114 |
Example 5 (2004 Singapore Math Olympiad) Let $0<a, b, c<1$, and $ab+bc+ca=1$. | Prove: $\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}}+\frac{c}{1-c^{2}} \geqslant \frac{3 \sqrt{3}}{2}$.
Proof: Let $f(x)=\frac{x}{1-x^{2}}$ and $0<x<1$, then $f^{\prime}(x)=\frac{1+x^{2}}{\left(1-x^{2}\right)^{2}}>0, f^{\prime \prime}(x)=\frac{2 x\left(3+x^{2}\right)}{\left(1-x^{2}\right)^{3}}>0$, thus $f(x)$ is a convex and in... | \frac{3 \sqrt{3}}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,115 |
Example 6 (2006 China Mathematical Olympiad Problem) The sequence of real numbers $\left\{a_{n}\right\}$ satisfies $a_{t}=\frac{1}{2}, a_{k+1}=-a_{k}+\frac{1}{2-a_{k}}, k=$ $1,2, \cdots, n$.
$$\begin{array}{c}
\text { Prove: } \quad\left[\frac{n}{2\left(a_{1}+a_{2}+\cdots+a_{n}\right)}-1\right]^{\prime \prime} \leqslan... | Prove: (1) Let the function $f(x)=-x+\frac{1}{2-x}, x \in \left[0, \frac{1}{2}\right]$, by mathematical induction we can get $0<a_{n} \leqslant \frac{1}{2}$ and by the problem statement we can prove $\left(\frac{n}{\sum_{k=1}^{n} a_{k}}\right)^{n}\left(\frac{n}{2 \sum_{k=1}^{n} a_{k}}-1\right)^{n} \leqslant \prod_{k=1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,116 |
\begin{array}{l}\text { Example 1 } \quad \text { Let } x_{k}>0(k=1,2, \cdots, n), x_{1}+x_{2}+\cdots \\ +x_{n}=1, p \in \mathbf{N}^{*}, p \geqslant 2 . \\ \text { Prove: } x_{1}^{p}+x_{2}^{+}+\cdots+x_{n}^{p} \geqslant n^{1-p} .\end{array} | Proof: Let the function $f(x)=x^{p}(0<x<1, p>1)$, then $f''(x)=p(p-1)x^{p-2}>0$, so $f(x)$ is a convex function on $(0,1)$.
By Jensen's inequality $f\left(\frac{1}{n} \sum_{i=1}^{n} x_{i}\right) \leqslant \frac{1}{n} \sum_{i=1}^{n} f\left(x_{i}\right)$, i.e., $f\left(\frac{1}{n}\right) \leqslant \frac{1}{n}\left(x_{1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,119 |
Example 2 Let $a_{1}, a_{2}, \cdots, a_{n}>0, n \geqslant 2$, and $a_{1}+a_{2}+$ $\cdots+a_{n}=1$.
Prove: $\frac{a_{1}}{2-a_{1}}+\frac{a_{2}}{2-a_{2}}+\cdots+\frac{a_{n}}{2-a_{n}} \geqslant$
$$\frac{n}{2 n-1} .$$ | Proof: Let $f(x)=\frac{x}{2-x}(0<x<1)$, then $f^{\prime}(x)=\frac{2}{(2-x)^{2}}>0$, and $f^{\prime \prime}(x)=\frac{4}{(2-x)^{3}}>0$, so $f(x)$ is a convex function on $(0,1)$.
By Jensen's inequality $f\left(\frac{1}{n} \sum_{i=1}^{n} a_{i}\right) \leqslant \frac{1}{n} \sum_{i=1}^{n} f\left(a_{i}\right)$, that is, $f\... | \frac{a_{1}}{2-a_{1}}+\frac{a_{2}}{2-a_{2}}+\cdots+\frac{a_{n}}{2-a_{n}} \geqslant \frac{n}{2 n-1} | Inequalities | proof | Yes | Yes | inequalities | false | 737,120 |
Example 3 If $a_{i}>0(i=1,2, \cdots, n, n \geqslant 2)$, and $\sum_{i=1}^{n} a_{i}=s(s$ is a positive number $)$.
Prove: $\sum_{i=1}^{n} \frac{a_{i}^{k}}{s-a_{i}} \geqslant \frac{s^{k-1}}{(n-1) n^{k-2}}$ (where $k$ is a constant, and $k$ is a positive integer). | Proof: Let $f(x)=\frac{x^{k}}{s-x}(0 < x < s, k > 1)$. Since $f''(x) > 0$, $f(x)$ is a convex function on $(0, s)$.
By Jensen's inequality, $f\left(\frac{1}{n} \sum_{i=1}^{n} a_{i}\right) \leqslant \frac{1}{n} \sum_{i=1}^{n} f\left(a_{i}\right)$, which means $f\left(\frac{s}{n}\right) \leqslant \frac{1}{n} \sum_{i=1}^... | \sum_{i=1}^{n} \frac{a_{i}^{k}}{s-a_{i}} \geqslant \frac{s^{k-1}}{(n-1) n^{k-2}} | Inequalities | proof | Yes | Yes | inequalities | false | 737,121 |
Example 5 (47th Polish Mathematical Olympiad, Second Round) Let $a, b, c > 0$, and $a + b + c = 1$.
Prove: $\frac{a}{a^{2}+1}+\frac{b}{b^{2}+1}+\frac{c}{c^{2}+1} \leqslant \frac{9}{10}$. | Proof: Let $f(x)=\frac{x}{x^{2}+1}(0<x<1)$, so $f^{\prime}(x)=$
$$\frac{1-x^{2}}{\left(x^{2}+1\right)^{2}} f^{\prime \prime}(x)=\frac{2 x\left(x^{2}-3\right)}{\left(x^{2}+1\right)^{3}},$$
Since $0<x<1$, we have $f^{\prime \prime}(x)<0$, so $f(x)$ is a concave function on $(0,1)$.
By Jensen's inequality, we get $\left... | \frac{9}{10} | Inequalities | proof | Yes | Yes | inequalities | false | 737,122 |
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