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Example 4 Given $a \geqslant b \geqslant c \geqslant 0$, and $a+b+c=$
3. Prove: $a b^{2}+b c^{2}+c a^{2} \leqslant \frac{27}{8}$. | Prove: When $a=b=\frac{3}{2}, c=0$, the equality of the inequality holds. Fix $b$, then $c=3-b-a$. He considers the function $f(a)=a b^{2}+b c^{2}+c a^{2}$, and $f(a)$ is decreasing in $c$ $f^{\prime}(a) \leqslant 0$ I initially wanted to adjust $c$ to $\Leftrightarrow b^{2}-2 b c+2 a c-a^{2} \leqslant 0$ up to $a+c$, ... | \frac{27}{8} | Inequalities | proof | Yes | Yes | inequalities | false | 737,722 |
Example 5 Given $x, y, z \in \mathbf{R}_{+} \cup\{0\}$, and $x+y+z=\frac{1}{2}$. Find
$$\frac{\sqrt{x}}{4 x+1}+\frac{\sqrt{y}}{4 y+1}+\frac{\sqrt{z}}{4 z+1}$$
the maximum value. | Solution: It's easy to guess that when $x=y=z=\frac{1}{6}$, equation (1) takes the maximum value $\frac{3}{5} \sqrt{\frac{3}{2}}$ (at least this is a "candidate" for the maximum value. Another candidate is when $x=\frac{1}{2}, y=z=0$, the function value is $\frac{1}{3} \sqrt{\frac{1}{2}}$. These special values are wort... | \frac{3}{5} \sqrt{\frac{3}{2}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,723 |
Example 6 Given $x, y, z \in \mathbf{R}_{+}$, and $x+y+z=1$. Find
$$\frac{\sqrt{x}}{4 x+1}+\frac{\sqrt{y}}{4 y+1}+\frac{\sqrt{z}}{4 z+1}$$
the maximum value. | Solution: The difference between this problem and Example 5 is that the condition $x+y+z=\frac{1}{2}$ is changed to $x+y+z=1$.
Assume $x \geqslant \frac{1}{3} \geqslant z$. Fix $y$, and let $f(x)=\sum \frac{\sqrt{x}}{4 x+1}$, which is strictly decreasing $\Leftrightarrow^{\prime}(x)0, x \in\left[\frac{1}{4}, \frac{1}{... | \frac{3 \sqrt{3}}{7} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,724 |
Example 7 Given $x, y, z \in \mathbf{R}_{+}$, and $xyz=1$. Prove:
$$\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}} \leqslant \frac{3 \sqrt{2}}{2} .$$ | Proof: Without loss of generality, let $x \geqslant y \geqslant z$.
When $y \leqslant 2$, fix $x$, then $z=\frac{1}{x y}$. The function
$$\begin{array}{l}
f(y)=\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}} \text{ is decreasing } \\
\Leftrightarrow f'(y) \leqslant 0 \\
\Leftrightarrow-\frac{1}{(1+y)^{\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,725 |
Example 9 Given a positive integer $n$. Find the smallest positive number $\lambda$, such that for any $\theta_{i} \in\left(0, \frac{\pi}{2}\right)(i=1,2, \cdots, n)$, as long as
$$\tan \theta_{1} \tan \theta_{2} \cdots \tan \theta_{n}=2^{\frac{n}{2}}$$
then
$$\cos \theta_{1}+\cos \theta_{2}+\cdots+\cos \theta_{n} \le... | When $n=1$,
$$\cos \theta_{1}=\left(1+\tan ^{2} \theta_{1}\right)^{-\frac{1}{2}}=\frac{\sqrt{3}}{3}, \lambda=\frac{\sqrt{3}}{3}$$
Assume $n \geqslant 2$. Let $x_{i}=\tan ^{2} \theta_{i}\left(1 \leqslant_{i} \leqslant_{n}\right)$, then the given condition becomes
$$x_{1} x_{2} \cdots x_{n}=2^{n}$$
We need to find the ... | \lambda = \frac{2}{\sqrt{3}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,727 |
Example 3 Let $A, B, C$ be the three interior angles of an acute $\triangle ABC$. Prove: $\sum \frac{\cos ^{2} A}{1+\cos A} \geqslant \frac{1}{2}$. | Proof: Let \( S(A) = \frac{\cos^2 A}{1 + \cos A} \)
\[ = \frac{1}{1 + \cos A} + \cos A - 1, \]
where \( 0 < A < \frac{\pi}{2} \). If \( S''(A) \) has a unique zero point \( \varphi \) satisfying \( \frac{\pi}{4} < \varphi < \frac{\pi}{2} \), then \( S''(A) \) has a unique zero point \( \varphi \) such that \( \frac{\p... | \frac{1}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 737,730 |
Proposition 1: From the focus of an ellipse, draw a perpendicular to any tangent line, the locus of the foot of the perpendicular is a circle. | Prove: Extend $F_{1} P$ to $T$, by Lemma 1, the tangent is the angle bisector of $\angle T P F_{2}$, draw $F_{2} R$ perpendicular to the tangent at $R$, extend $F_{2} R$ to intersect $P T$ at $S$, then in the isosceles $\triangle P F_{2} S$, $\left|P F_{2}\right|$ $=|P S|$, in $\triangle S F_{1} F_{2}$,
$\therefore$ b... | x^{2}+y^{2}=a^{2} | Geometry | proof | Yes | Yes | inequalities | false | 737,731 |
Proposition 2: The locus of the foot of the perpendicular drawn from the focus of a hyperbola to its tangent is a circle. | Prove: Draw $F_{2} R$ perpendicular to the tangent at $R$, extend it to intersect $P F_{1}$ at $S$. By Lemma 2, $P Q$ is the angle bisector of $\angle F_{1} P F_{2}$, so $\triangle S P F_{2}$ is an isosceles triangle, hence $|S P| = |P F_{2}|$,
By the definition of a hyperbola: $2a = |P F_{1}| - |P F_{2}| = |P F_{1}| ... | x^{2} + y^{2} = a^{2} | Geometry | proof | Yes | Yes | inequalities | false | 737,732 |
Proposition 3: The locus of the foot of the perpendicular drawn from the focus of a parabola to its tangent is the straight line passing through the vertex of the parabola and perpendicular to the axis. | Prove: The equation of the tangent line at point $P\left(x_{0}, y_{0}\right)$ on the parabola is
$y_{0} y=p\left(x+x_{0}\right)$, i.e., $y_{0} y=p x+\frac{y_{0}^{2}}{2} \cdots$ (1)
Draw a line perpendicular to the tangent line through $F\left(\frac{p}{2}, 0\right)$, its equation is $y=-\frac{y_{0}}{p}\left(x-\frac{p}{2... | proof | Geometry | proof | Yes | Yes | inequalities | false | 737,733 |
Theorem: If $a, b$ are positive numbers, and the real number $\lambda \geqslant 3$, then $\frac{1}{\sqrt{\lambda}}<\frac{1}{\sqrt{\lambda+a}}+\frac{1}{\sqrt{\lambda+b}} \leqslant \frac{2}{\sqrt{\lambda+1}} \leqslant \frac{1}{\sqrt{1+\lambda a}}+\frac{1}{\sqrt{1+\lambda b}}$.
For $a, b > 0$, and $x y=1$, for $\lambda \... | Proof: For $x>0$, consider the function $f(x)=\frac{1}{\sqrt{1+\lambda x}}+\frac{1}{\sqrt{1+\frac{\lambda}{x}}}$ with parameter $\lambda \in\left(0, \frac{1}{3}\right]$. Taking the derivative, we get $f^{\prime}(x)=-\frac{\lambda}{2}(1+\lambda x)^{-\frac{3}{2}}+\frac{\lambda}{2 x^{2}}\left(1+\frac{\lambda}{x}\right)^{-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,734 |
Let there be two different media I and II on either side of the $x$-axis (Figure 2), with the speed of light in media I and II being $c_{1}$ and $c_{2}$, respectively. If a light ray starts from point A in medium I and reaches point B in medium II, what path should the light take to minimize the time of travel? | Solution: In the same medium, the shortest path of a light ray is a straight line. We can assume that the light ray travels along the straight line AP in medium I and along the straight line PB in medium II, using the notation in the figure. Then,
$$\mathrm{AP}=\sqrt{x^{2}+b_{1}^{2}}, \mathrm{~PB}=\sqrt{(a-x)^{2}+b_{2}... | \frac{\sin \theta_{1}}{\sin \theta_{2}}=\frac{c_{1}}{c_{2}} | Calculus | math-word-problem | Yes | Yes | inequalities | false | 737,737 |
Example 14 Let $x, y, z \geqslant 0, xy + yz + zx = 1$. Then
$$f(x, y, z) = \frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} \geqslant \frac{5}{2}$$ | Explanation: Suppose $x \geqslant y \geqslant z$, and let $u=\frac{x+y}{2} \geqslant$ $z$. From the given conditions, we have $u^{2} \geqslant \frac{1}{3}$.
When $u \geqslant 1$,
$$f \geqslant \frac{1}{x+y}+x+y=2 u+\frac{1}{2 u} \geqslant \frac{5}{2}$$
When $u < 1,$
Thus, it suffices to prove
$$f(x, y, z) \geqslant ... | \frac{5}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 737,742 |
Example 15 Let $S=\frac{1}{\sqrt[3]{4}}+\cdots+\frac{1}{\sqrt[3]{1000}}$. Prove that $146<S<147$ | Explain: Using the method of partial fraction summation, first prove
$$\begin{array}{l}
\frac{3}{2}\left(\sqrt[3]{(k+1)^{2}}-\sqrt[3]{k^{2}}\right)<\frac{1}{\sqrt[3]{k}} \\
<\frac{3}{2}\left(\sqrt[3]{k^{2}}-\sqrt[3]{(k-1)^{2}}\right)
\end{array}$$
The terms of the partial fraction are precisely the antiderivative of t... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,743 |
Example 16 Given a positive integer $n(n \geqslant 2)$, let positive integers $a_{i}(i=1,2, \cdots, n)$ satisfy
$$a_{1}<a_{2}<\cdots<a_{n} \text { and } \sum_{i=1}^{n} \frac{1}{a_{i}} \leqslant 1 \text {. }$$
Prove that for any $x$, we have
$$\left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} \leqslant \frac{1}... | Explanation: Since the square of the sum of $n$ terms on the left side of equation (1) is no more than the single term on the right side, this leads us to think that we must simplify the sum on the left side of equation (1) so that only one term remains in the end. The most commonly used method for summing non-special ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,744 |
Example 1 For all positive real numbers $a, b, c, d$, prove:
$$\begin{array}{l}
\frac{a}{b+2 c+3 d}+\frac{b}{c+2 d+3 a}+\frac{c}{d+2 a+3 b}+ \\
\frac{d}{a+2 b+3 c} \geqslant \frac{2}{3} .
\end{array}$$ | Prove: Perform the linear transformation
$$\left\{\begin{array}{l}
x=b+2 c+3 d, \\
y=c+2 d+3 a, \\
z=d+2 a+3 b, \\
w=a+2 b+3 c .
\end{array}\right.$$
Considering \(a, b, c, d\) as variables, solving the system of equations yields
$$\left\{\begin{array}{l}
a=-\frac{5}{24} x+\frac{7}{24} y+\frac{1}{24} z+\frac{1}{24} w ... | \frac{2}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 737,749 |
Example 2 If $x, y, z > 1$, and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$, prove:
$$\sqrt{x+y+z} \geqslant \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} .$$ | Proof: Let $\alpha, \beta, \gamma$ all be acute angles, then $\frac{1}{\cos ^{2} \alpha}>1, \frac{1}{\cos ^{2} \beta}>1, \frac{1}{\cos ^{2} \gamma}>1$.
Make the trigonometric substitution
$$x=\frac{1}{\cos ^{2} \alpha}, y=\frac{1}{\cos ^{2} \beta}, z=\frac{1}{\cos ^{2} \gamma} .$$
Then the condition simplifies to
$$\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,750 |
Example 3 Let $a, b, c$ be positive numbers, and $abc=1$. Try to prove:
$$\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2} \text {. }$$ | To prove Example 3, we first provide a generalization of Example 3.
Let \( m \in \mathbf{N}, m \geqslant 2, a, b, c \in \mathbf{R}_{+} \) and \( abc = 1 \). Try to prove:
$$
\frac{1}{a^{m}(b+c)} + \frac{1}{b^{m}(c+a)} + \frac{1}{c^{m}(a+b)} \geqslant \frac{3}{2}
$$
Analysis: Polya said, solving a problem is to turn th... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,751 |
Example 1. Let $x_{1}, x_{2}, \cdots, x_{n}$ be angles between $\mathcal{O}$ and $180^{\circ}$, then
$$\begin{array}{l}
\sin x_{1}+\sin x_{2}+\cdots+\sin x_{n} \leqslant n \sin \left(\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) \\
\sin x_{1} \sin x_{2} \cdots \sin x_{n} \leqslant\left(\sin \frac{x_{1}+x_{2}+\cdots+x_{n}}{... | Prove that $\sin x$ is always positive on $\left(0^{\circ}, 180^{\circ}\right)$. If $x_{1}, x_{2}, y_{1}$, $y_{2} \in\left(0^{\circ}, 180^{\circ}\right)$, and $x_{1}+x_{2}=y_{1}+y_{2}$, then
$$\begin{array}{l}
\sin x_{1}+\sin x_{2}<\sin y_{1}+\sin y_{2} \Leftrightarrow \cos \frac{x_{1}-x_{2}}{2}<\cos \frac{y_{1}-y_{2}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,753 |
Example 2. In any $\triangle ABC$, the following inequalities hold:
(1) $\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \leqslant \frac{1}{8}$
(2) $\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \leqslant \frac{3 \sqrt{3}}{8}$
(3) $\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2} \leqslant \frac{3}{2}$
(4) $\... | Proof: The proofs of these inequalities are largely similar, and here we only prove (10). We have
$$\begin{array}{l}
\operatorname{tg}^{2} \frac{A}{2}+\operatorname{tg}^{2} \frac{B}{2}=\frac{4\left(\sin \frac{A+B}{2}\right)^{2}-2 \sin A \sin B}{\left(\cos \frac{A-B}{2}+\cos \frac{A+B}{2}\right)^{2}} \\
=\frac{-2\left(\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,754 |
Example 1 Prove that $\log _{2} \pi+\log _{5} \pi>2$. | Just change $\lg 2$ to $\frac{1}{2} \lg 4$, and let $\lg 4$ participate in the inequality's expansion and contraction.
$$\begin{array}{l}
\quad \text { Proof Left }=\lg \pi \cdot \frac{1}{\lg 2 \cdot \lg 5}=\lg \pi \cdot \frac{2}{\lg 4 \cdot \lg 5}> \\
2 \lg \pi\left(\frac{2}{\lg 4+\lg 5}\right)^{2}=8 \lg \pi\left(\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,755 |
Example 2 If $a+b+c=1, a, b, c \in R^{+}$, prove that $\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right) \geqslant 64$.
untranslated text remains unchanged. | Analyzing, one generally thinks that from $a+b+c=1$ we get $a b c \leqslant \frac{1}{27}$.
$$\begin{array}{l}
(a+1)(b+1)(c+1)-64 a b c=a b+b c+c a+a+b+c \\
+1-63 a b c \geqslant a b+b c+c a+2-\frac{63}{27}=a b+b c+c a-
\end{array}$$
$\frac{1}{3}$
However, under the given conditions, it is easy to see that $a b+b c+c a ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,756 |
Example 3 Prove: $1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{n^{2}}<\frac{7}{4}$. | $$
\because \frac{1}{k^{2}}=\frac{1}{k \cdot k}1) (*), \therefore \text{left side } 1)
\end{array}$$
When $n$ is even: $\frac{1}{2^{2}}+\frac{1}{4^{2}}+\cdots+\frac{1}{n^{2}}<\frac{1}{2}[(1-$
$$\begin{array}{c}
\left.\left.\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots+\left(\frac{1}{n-1}-\frac{1}{n+1}... | \frac{7}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 737,757 |
Example $1: x>0$, prove that: $x-\frac{x^{2}}{2}-\ln (1+x)<0$ | Prove: Let $f(x)=x-\frac{x^{2}}{2}-\ln (1+x)(x>0)$, then $f^{\prime}(x)=-\frac{x^{2}}{1+x}$ $\because x>0, \therefore f^{\prime}(x)<0$ when $x>0$, $f(x)<f(0)=0$, i.e., $x-\frac{x^{2}}{2}-\ln (1+x)<0$ holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,762 |
Example 2: When $x \in(0, \pi)$, prove the inequality $\sin x < x$ holds.
untranslated text:
当 $x \in(0, \pi)$ 时,证明不等式 $\sin x<x$ 成立。
translated text:
When $x \in(0, \pi)$, prove the inequality $\sin x < x$ holds. | Prove: Let $f(x)=\sin x-x$, then $f^{\prime}(x)=\cos x-1$.
$\because x \in(0, \pi), \therefore f^{\prime}(x)<0$. $\therefore f(x)=\sin x-x$ is monotonically decreasing in $x \in(0, \pi)$, and $f(0)=0$. $\square$
$\therefore f(x)=\sin x-x<f(0)=0$, hence when $x \in(0, \pi)$, $\sin x<x$ holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,763 |
Example 3: Prove that: $\frac{1}{3}(x-1)>x^{\frac{1}{3}}-1$, where $x \in(1,+\infty)$. | Prove: Let $f(x)=\frac{1}{3}(x-1)-x^{\frac{1}{3}}+1$, then $f^{\prime}(x)=\frac{1}{3}-\frac{1}{3} x^{-\frac{2}{3}}=\frac{1}{3}(1-$ $\left.\frac{1}{\sqrt[3]{x^{2}}}\right)$
$\because x>1, \therefore \frac{1}{\sqrt[3]{x^{2}}}<1$, thus $\frac{1}{3}(1-\frac{1}{\sqrt[3]{x^{2}}})>0$, i.e., $f^{\prime}(x)>0$, so $f(x)$ is an ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,764 |
Example 4: Prove that for $n \in N^{*}, n \geqslant 3$, $2^{n}>2 n+1$ | To prove the original statement, we need to prove: $2^{n}-2 n-1>0, n \geqslant 3$ holds.
Let $f(x)=2^{x}-2 x-1(x \geqslant 3)$, then $f'(x)=2^{x} \ln 2-2(x \geqslant 3)$,
$\because x \geqslant 3, \therefore f'(x) \geqslant 2^{3} \ln 2-2>0 . \therefore f(x)$ is an increasing function on $[3,+\infty)$,
$\therefore$ the m... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,765 |
Example 5: Prove the inequality: $x^{2 a}-2 a x \leq 1-a \ (x>0, a>1)$ | Proof: Let $f(x)=x^{2a}-2ax-(1-a)$, then $f^{\prime}(x)=2ax^{2a-1}-2a(x^{2a-1}-1)$
Let $f^{\prime}(x)=0$, we get the unique critical point $x=1$
Also, when $0<x<1$, $f^{\prime}(x)<0$; when $x>1$, $f^{\prime}(x)>0, (a>0, a>1)$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,766 |
Example $6: f(x)=\frac{1}{3} x^{3}-x, x_{1}, x_{2} \in[-1,1]$, prove that: $\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| \leqslant \frac{4}{3}$ | Proof: $\because f^{\prime}(x)=x^{2}-1, x \in[-1,1]$ when, $f^{\prime}(x) \leqslant 0$,
$\therefore f(x)$ is decreasing on $[-1,1]$. Therefore, the maximum value of $f(x)$ on $[-1,1]$ is $f(-1)$ $=\frac{2}{3}$
the minimum value is $f(1)=-\frac{2}{3}$, i.e., the range of $f(x)$ on $[-1,1]$ is $\left[-\frac{2}{3}, \frac... | \frac{4}{3} | Calculus | proof | Yes | Yes | inequalities | false | 737,767 |
Example 7. Given the function $f(x)=a x^{2}+\frac{1}{x}-2 \ln x(x>0)$. If $f(x)$ is monotonically increasing on $[1,+\infty)$, find the range of the real number a; | Solution: Given $f(x)=a x^{2}+\frac{1}{x}-2 \ln x$, we get $f^{\prime}(x)=2 a x-\frac{1}{x^{2}}-\frac{2}{x}$.
Since the function $f(x)$ is a monotonically increasing function on $[1,+\infty)$, we have $f^{\prime}(x) \geq 0$ on $[1,+\infty)$,
which means the inequality $2 a x-\frac{1}{x^{2}}-\frac{2}{x} \geq 0$ holds f... | a \geq \frac{3}{2} | Calculus | math-word-problem | Yes | Yes | inequalities | false | 737,768 |
Example 1 Given $a+b+c=1$. Prove: $a^{2}+b^{2}+c^{2} \geqslant \frac{1}{3}$. | Analysis: The terms on the left side of the inequality to be proven are all quadratic, while the constant $\frac{1}{3}$ on the right side is of zero degree. Therefore, it is believed that the degrees of the structures on both sides of the inequality are unbalanced, so it is particularly important to convert the right s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,769 |
Example 2 Given $a+b+c=1$, and $a, b, c$ are all non-negative real numbers. Prove: $\sqrt{a}+\sqrt{b}+\sqrt{c} \leqslant \sqrt{3}$. | Analysis: This problem has the same condition as Example 1, which is $a+b+c=1$. The terms on the left side of the inequality to be proven are all $\frac{1}{2}$ in power, while the constant $\sqrt{3}$ on the right side is of zero power. Therefore, $\sqrt{a+b+c}=1$ should be substituted into the right side, transforming ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,770 |
Example 11 Let $x_{i}>0(i=1,2,3, \cdots, n)$, prove:
$$\frac{x_{1}^{2}}{x_{2}}+\frac{x_{2}^{2}}{x_{3}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}}+\frac{x_{n}^{2}}{x_{1}} \geqslant x_{1}+x_{2}+\cdots+x_{n} .$$ | Analysis: The inequality to be proved is equivalent to $\frac{x_{1}^{2}}{x_{2}}-x_{1}+\frac{x_{2}^{2}}{x_{3}}-x_{2}+\cdots$ $+\frac{x_{n}^{2}}{x_{1}}-x_{n} \geqslant 0$, and when $x_{i}>0$, for any $x_{j}, x_{k}(1 \leqslant j, k$ $\leqslant n)$, we always have $\frac{x_{k}}{x_{j}}\left(x_{k}-x_{j}\right) \geqslant x_{k... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,771 |
Example 12 If the three sides of a triangle are $a, b, c$. Prove:
$$(a+b+c)^{3} \geqslant 27(a+b-c)(b+c-a)(c+a-b) .$$ | Analysis: Considering $a, b, c$ are the three sides of a triangle, and the sum of any two sides is greater than the third side, we can set $a+b-c=x, b+c-a=y$, $c+a-b=z$, then we have $x>0, y>0, z>0$, and $x+y+z=a$ $+b+c$. Also, $\frac{x+y+z}{3} \geqslant \sqrt[3]{x y z}$, thus $\frac{(x+y+z)^{3}}{27} \geqslant x y z$, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,772 |
Example 13 If $x+y+z=a$, and $x, y, z \in \mathrm{R}$. Prove:
$$x^{2}+y^{2}+z^{2} \geqslant \frac{a^{2}}{3 .}$$ | Analysis: Due to the "equal status" of $x$, $y$, and $z$, we can use the relationship between $x$, $y$, and $z$ and their average value $\frac{a^{2}}{3}$ to reduce the number of variables through substitution. Let $x=\frac{a}{3}+\alpha, y=\frac{a}{3}+\beta, z=\frac{a}{3}-(\alpha+\beta)$, $\alpha, \beta \in \mathrm{R}$,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,773 |
Example 14 Given that $a$ and $b$ are both positive numbers, and $a+b=1$. Prove:
$$\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2} \geqslant \frac{25}{2} .$$ | Analysis: Given the structure of $a, b$ being positive numbers and $a+b=1$, we are reminded of the trigonometric identity $\sin ^{2} \alpha+\cos ^{2} \alpha=1$. Therefore, we can make a trigonometric substitution. Let $a=\sin ^{2} \alpha, b=\cos ^{2} \alpha,\left(0<\alpha<\frac{\pi}{2}\right)$, then
$$\begin{array}{l}
... | \frac{25}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 737,774 |
Example 15 Given that $a$, $b$, and $c$ are all positive numbers, and $a^{2}+b^{2}=c^{2}$. Prove: $a^{n}+b^{n}<c^{n}$. | Analysis: Given that $a$, $b$, and $c$ are all positive numbers and $a^{2}+b^{2}=c^{2}$, it follows that $a$, $b$, and $c$ can be the lengths of the sides of a right triangle. Therefore, we can let $a=c \cos \alpha, b=c \sin \alpha, (0<\alpha<\frac{\pi}{2})$, so $a^{n}+b^{n}=c^{n}(\sin ^{n} \alpha+\cos ^{n} \alpha)<c^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,775 |
Example 16 Given that $a, b, c$ are all positive numbers. Prove:
$$\frac{a^{4}}{b^{2}+c^{2}}+\frac{b^{4}}{c^{2}+a^{2}}+\frac{c^{4}}{a^{2}+b^{2}} \geqslant \frac{a^{2}+b^{2}+c^{2}}{2} .$$ | Analysis: Since the equality holds when $a=b=c$, at this time $\frac{a^{4}}{b^{2}+c^{2}}$
$$\begin{array}{l}
=\frac{b^{2}+c^{2}}{4}, \frac{b^{4}}{c^{2}+a^{2}}=\frac{c^{2}+a^{2}}{4}, \frac{c^{4}}{a^{2}+b^{2}}=\frac{a^{2}+b^{2}}{4}, \text { and } \\
\frac{a^{4}}{b^{2}+c^{2}}+\frac{b^{2}+c^{2}}{4} \geqslant a^{2}, \frac{b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,776 |
Example 17 Given that $a, b, c$ are all positive numbers, and satisfy $abc=1$.
Prove: $\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2}$. | Analysis: The original inequality is equivalent to $\frac{b^{2} c^{2}}{a(b+c)}+\frac{c^{2} a^{2}}{b(c+a)}$ $+\frac{a^{2} b^{2}}{c(a+b)} \geqslant \frac{3}{2}$. When $a=b=c=1$, equality holds, at which point $\frac{b^{2} c^{2}}{a(b+c)}=\frac{a(b+c)}{4}, \frac{c^{2} a^{2}}{b(c+a)}=\frac{b(c+a)}{4}, \frac{a^{2} b^{2}}{c(a... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 737,777 |
Example 18 Given that $\alpha, \beta, \gamma$ are all acute angles, and satisfy
$$\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$$
Prove: $\frac{1}{\sin ^{2} \alpha}+\frac{1}{\sin ^{2} \beta}+\frac{1}{\sin ^{2} \gamma} \geqslant \frac{9}{2}$. | Analysis: Since the equality holds when $\sin ^{2} \alpha=\sin ^{2} \beta=\sin ^{2} \gamma=\frac{2}{3}$, we have $\frac{1}{\sin ^{2} \alpha}+\frac{9 \sin ^{2} \alpha}{4} \geqslant 3, \frac{1}{\sin ^{2} \beta}+\frac{9 \sin ^{2} \beta}{4} \geqslant 3, \frac{1}{\sin ^{2} \gamma}+$ $\frac{9 \sin ^{2} \gamma}{4} \geqslant 3... | \frac{1}{\sin ^{2} \alpha}+\frac{1}{\sin ^{2} \beta}+\frac{1}{\sin ^{2} \gamma} \geqslant \frac{9}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 737,778 |
Example 3 Given $a>0, b>0$, and $a^{3}+b^{3}=2$. Prove: $a+b \leqslant 2$ | Analysis: The given expression is a cubic polynomial, and the expression to be proven is a linear polynomial. It is difficult to reduce the cubic polynomial to a linear one, but it is easy to elevate the linear polynomial to a cubic one. Therefore, the expression to be proven is that \((a+b)^{3} \leqslant 8\) holds. Af... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,779 |
Example 4 Given that $a, b, c$ are all positive numbers. Prove:
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9 \text {. }$$ | Analysis: The left side of the equation to be proven is the product of two trinomials, while the right side is the constant 9. If we use the arithmetic mean inequality once, $a, b, c \in \mathbb{R}^{+} \Rightarrow a+b+c \geqslant 3 \sqrt[3]{a b c}$, a coefficient 3 will appear, so the constant 9 on the right side $=3 \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,780 |
Example 6 Given that $a, b, c$ are all positive numbers, prove:
$$\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a} \geqslant a+b+c$$ | Analysis: From the "degree" and "number of terms" on both sides of the expression to be proved, there is no need to balance. From the structure of the operations, the left side of the expression to be proved is a fraction, while the right side is a polynomial, so removing the denominator becomes the key to solving the ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,782 |
Example 7 Given that $a, b, c$ are distinct positive real numbers, and $abc = 1$. Prove: $\sqrt{a}+\sqrt{b}+\sqrt{c}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. | Analysis: Since the right side of the inequality to be proved is a fraction, we can use the condition $a b c=1$ to transform the left side into a fraction:
$$\begin{aligned}
\sqrt{a}+\sqrt{b}+\sqrt{c} & =\sqrt{\frac{1}{b c}}+\sqrt{\frac{1}{a c}}+\sqrt{\frac{1}{a b}} \\
& <\frac{1}{2}\left(\frac{1}{b}+\frac{1}{c}\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,783 |
Example 8 Given $a, b, c \in \mathrm{R}^{+}$. Prove:
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2} .$$ | Analysis: Distributing the constant $\frac{3}{2}$ uniformly to each term on the left side, the inequality to be proven is equivalent to $\frac{a}{b+c}-\frac{1}{2}+\frac{b}{c+a}-\frac{1}{2}+\frac{c}{a+b}-\frac{1}{2} \geqslant 0$ (*). By the symmetry of $a, b, c$, without loss of generality, assume $a \geqslant b \geqsla... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,784 |
Example 9 In $\triangle ABC$, prove that:
$$\cos A+\cos B+\cos C \leqslant \frac{3}{2} .$$ | Analysis: The inequality to be proved is equivalent to $\cos A-\frac{1}{2}+\cos B-\frac{1}{2}$ $+\cos C-\frac{1}{2} \leqslant 0$, i.e., $1-2 \cos A+1-2 \cos B+1-2 \cos C$ $\geqslant 0$. Since $C=\pi-(A+B)$, then
$$\cos C=-\cos (A+B)=-\cos A \cos B+\sin A \sin B,$$
Thus, $1-2 \cos A+1-2 \cos B+1-2 \cos C$
$$\begin{alig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,785 |
Example 10 Given that $a, b, c$ are all positive numbers. Prove:
$$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{a+b+c}{2} .$$ | Analysis: Decompose and reorganize the independent term $\frac{a+b+c}{2}$, then the inequality to be proved is equivalent to $\frac{a^{2}}{b+c}-\frac{a}{2}+\frac{b^{2}}{c+a}-\frac{b}{2}+\frac{c^{2}}{a+b}-\frac{c}{2} \geqslant 0$. By the symmetry of $a, b, c$, without loss of generality, assume $a \geqslant b \geqslant ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,786 |
Example 1 (Problem 2 of the 42nd IMO in 2001) For all positive real numbers $a, b, c$, prove that
$$\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 c a}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant 1$$ | Proof: Let $p$ be a constant to be determined, such that
$$\frac{a}{\sqrt{a^{2}+8 b c}} \geqslant \frac{a^{p}}{a^{p}+b^{p}+c^{p}}$$
Since $(*) \Leftrightarrow\left(a^{p}+b^{p}+c^{p}\right)^{2} \geqslant a^{2 p-2}\left(a^{2}+8 b c\right) \Leftrightarrow$
$$2 a^{p}\left(b^{p}+c^{p}\right)+\left(b^{p}+c^{p}\right)^{2} \g... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,787 |
Example 2 (Problem 3 of the 46th IMO in 2005) Let $x, y, z$ be positive numbers and $x y z \geqslant 1$. Prove that:
$$\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}-y^{2}}{x^{2}+y^{5}+z^{2}}+\frac{z^{5}-z^{2}}{x^{2}+y^{2}+z^{5}} \geqslant 0 .$$ | Prove: First, prove the "part inequality":
$$\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}} \geqslant \frac{x^{2}-\frac{1}{2}\left(y^{2}+z^{2}\right)}{x^{2}+y^{2}+z^{2}}$$
Since \(x^{3}\left(x^{3}-1\right)\left(x^{2}+y^{2}+z^{2}\right)-\left(x^{3}-1\right)\left(x^{5}+y^{2}+z^{2}\right)=\left(x^{3}-1\right)^{2}\left(y^{2}+z^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,788 |
Example 11 (1996, 47th Polish Mathematical Olympiad, Second Round) Let $a, b, c \geqslant-\frac{3}{4}$, and $a+b+c=1$, prove that $\frac{a}{a^{2}+1}$ $+\frac{b}{b^{2}+1}+\frac{c}{c^{2}+1} \leqslant \frac{9}{10}$ | Prove: Let the undetermined constant $\lambda$, such that $\frac{a}{a^{2}+1}-\frac{3}{10} \leqslant$ $\lambda\left(a-\frac{1}{3}\right) \Leftrightarrow \frac{-(3 a-1)(a-3)}{10\left(a^{2}+1\right)} \leqslant \frac{\lambda}{3}(3 a-1)$, considering the equality holds, we should have $\frac{-(3 a-1)(a-3)}{10\left(a^{2}+1\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,789 |
Example 12 (2005 Moldova Selection Contest)
Let $a, b, c > 0$, and $a^{4}+b^{4}+c^{4}=3$, prove that $\frac{1}{4-ab}+\frac{1}{4-bc}+\frac{1}{4-ca}$ $\leqslant 1$ | Proof: From the given, we have $a^{4}$ $0,4-a^{2}>0,4-b^{2}>0$, because $\frac{1}{4-a b} \leqslant$ $\frac{1}{2}\left(\frac{1}{4-a^{2}}+\frac{1}{4-b^{2}}\right) \Leftrightarrow 2\left(4-a^{2}\right)\left(4-b^{2}\right) \leqslant(4-$ $a b)\left(8-a^{2}-b^{2}\right) \Leftrightarrow(a b+4)(a-b)^{2} \geqslant 0$, so, $\fra... | 1 | Inequalities | proof | Yes | Yes | inequalities | false | 737,790 |
Example 13 (2003 Hunan High School Mathematics Competition Question)
Let $x, y$, $z>0$ and $x+y+z=1$, find the minimum value of $f(x, y, z)=\frac{3 x^{2}-x}{1+x^{2}}+$ $\frac{3 y^{2}-y}{1+y^{2}}+\frac{3 z^{2}-z}{1+z^{2}}$. | Given $x, y, z > 0$ and $x + y + z = 1$, we know $0 < x, y, z \leq 1$. First, we prove:
$$\frac{3 x^{2} - x}{1 + x^{2}} \geq \frac{9}{10} x - \frac{3}{10}$$
Since $(*) \Leftrightarrow 10(3 x^{2} - x) \geq (9 x - 3)(x^{2} + 1) \Leftrightarrow 9 x^{3} - 33 x^{2} + 19 x - 3 \leq 0 \Leftrightarrow (3 x - 1)^{2}(x - 3) \le... | 0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,791 |
Example 14 (2004 Singapore Mathematical Olympiad)
Let $0 < a, b, c < 1, ab + bc + ca = 1$, prove that $\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}} +\frac{c}{1-c^{2}} \geqslant \frac{3 \sqrt{3}}{2}$. | Prove: $\because 0<a<1, \therefore 2 a^{2}\left(1-a^{2}\right)^{2}=2 a^{2}(1-$ $\left.a^{2}\right)\left(1-a^{2}\right) \leqslant\left(\frac{2}{3}\right)^{3}$, so $a\left(1-a^{2}\right) \leqslant \frac{2 \sqrt{3}}{9} \Leftrightarrow$ $\frac{a}{1-a^{2}} \geqslant \frac{3 \sqrt{3}}{2} a^{2}$, similarly, $\frac{b}{1-b^{2}}... | \frac{3 \sqrt{3}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 737,792 |
Example 15 (1998 39th IMO Shortlist)
Let $x, y, z > 0, xyz = 1$, prove that: $\frac{x^{3}}{(1+y)(1+z)}+$
$$\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)} \geqslant \frac{3}{4} .$$ | Proof: By the AM-GM inequality, we have
\[
\frac{x^{3}}{(1+y)(1+z)} + \frac{1+y}{8} + \frac{1+z}{8} \geqslant 3 \sqrt[3]{\frac{x^{3}}{(1+y)(1+z)} \cdot \frac{1+y}{8} \cdot \frac{1+z}{8}} = \frac{3}{4} x,
\]
so
\[
\frac{x^{3}}{(1+y)(1+z)} \geqslant \frac{1}{8}(6 x - y - z - 2).
\]
Similarly, we have
\[
\frac{y^{3}}{(1+z... | \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 737,793 |
Example 16 (36th IMO 1995) Let $a, b, c > 0$ and $abc = 1$, prove that: $\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}$ $\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2}$. | Proof: Let $t$ be a positive constant to be determined. By the AM-GM inequality, we have $\frac{1}{a^{3}(b+c)}+t a(b+c) \geqslant \frac{2 \sqrt{t}}{a}$, so $\frac{1}{a^{3}(b+c)} \geqslant \frac{2 \sqrt{t}}{a} - t \frac{(b+c)}{bc} = 2 \sqrt{t} \frac{1}{a} - t\left(\frac{1}{b} + \frac{1}{c}\right)$, with equality if and ... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 737,794 |
Example 17 (1990 31st IMO Shortlist) Let $a, b, c, d > 0$, and $ab + bc + cd + da = 1$. Prove that $\frac{a^{3}}{b+c+d} + \frac{b^{3}}{c+d+a} + \frac{c^{3}}{d+a+b} + \frac{d^{3}}{a+b+c} \geqslant \frac{1}{3}$. | Proof: Let $t$ be a positive constant to be determined. By the AM-GM inequality, we have
$$\frac{a^{2}}{b+c+d}+t^{2}(b+c+d) \geqslant 2 t a \text {, i.e., } \frac{a^{3}}{b+c+d} \geqslant 2 t a^{2}$$
$$-t^{2} a(b+c+d)$$, with equality if and only if $t=\frac{a}{b+c+d}$.
Similarly, we have $\frac{b^{3}}{c+d+a}=2 t b^{2}-... | \frac{1}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 737,795 |
Example 18 (2005 China Mathematical Olympiad National Training Team Selection Test Question) Let $a, b, c, d>0$, and $abcd=1$, prove that
$$\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geqslant 1 .$$ | Prove: First, prove
$$\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}} \geqslant \frac{1}{1+a b}$$
In fact, $(*) \Leftrightarrow(1+a b)\left[(1+a)^{2}+(1+b)^{2}\right]$ $\geqslant[(1+a)(1+b)]^{2} \Leftrightarrow 1+\left(a^{2}+b^{2}\right) a b-2 a b-$ $(a b)^{2} \geqslant 0$, and $1+\left(a^{2}+b^{2}\right) a b-2 a b-(a b)^{2} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,796 |
Example 19 (2004 33rd US Mathematical Olympiad) Let $a, b, c$ be positive real numbers, prove: $\left(a^{5}-a^{2}+3\right)\left(b^{5}-b^{2}+3\right)$ $\left(c^{5}-c^{2}+3\right) \geqslant(a+b+c)^{3}$. | Proof: First, we prove the "part inequality": $a^{5}-a^{2}+3 \geqslant a^{3}+2$. This is known from $\left(a^{5}-a^{2}+3\right)-\left(a^{3}+2\right)=(a-1)^{2}(a+1)\left(a^{2}+a+1\right) \geqslant 0$, so this inequality holds. Similarly, we have $b^{5}-b^{2}+3 \geqslant b^{3}+2, c^{5}-c^{2}+3 \geqslant c^{3}+2$. Multipl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,797 |
Example 20 Let $a_{i}>0\left(i=1,2, \cdots, n, \sum_{i=1}^{n} a_{i}=1, k \in \mathbf{N}\right)$, prove: $\left(1+a_{1}^{k}+\frac{1}{a_{1}^{k}}\right)\left(1+a_{2}^{k}+\frac{1}{a_{2}^{k}}\right) \cdots\left(1+a_{n}^{k}+\frac{1}{a_{n}^{k}}\right) \geqslant$ $\left(1+n^{k}+\frac{1}{n^{k}}\right)^{n}$ | Proof: By the AM-GM inequality, we have $1+a_{1}^{k}+\frac{1}{a_{1}^{k}}=$
$$\begin{array}{l}
\underbrace{\frac{1}{n^{k}}+\frac{1}{n^{k}}+\cdots+\frac{1}{n^{k}}}_{n^{k}}+a_{1}^{k}+\underbrace{\frac{1}{n^{2 k} a_{1}^{k}}+\frac{1}{n^{2 k} a_{1}^{k}}+\cdots+\frac{1}{n^{2 k} a_{1}^{k}}}_{n^{2 k}} \geqslant \\
\left(n^{2 k}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,798 |
Example 3 (2004 Beijing Mathematics Competition for High School Students, Re-test) Let $a, b, c<0$, prove:
$$\frac{a^{4}}{4 a^{4}+b^{4}+c^{4}}+\frac{b^{4}}{a^{4}+4 b^{4}+c^{4}}+\frac{a^{4}}{a^{4}+b^{4}+4 c^{2}} \leqslant \frac{1}{2} \text {. }$$ | Proof: By the AM-GM inequality, we have $2 a^{4}+2 a^{4}+b^{4}+c^{4} \geqslant$ $2 a^{4}+2 a^{2} b^{2}+2 a^{2} c^{2}$, so $\frac{a^{4}}{4 a^{4}+b^{4}+c^{4}} \leqslant$ $\frac{a^{4}}{2 a^{4}+2 a^{2} b^{2}+2 a^{2} c^{2}}=\frac{1}{2} \frac{a^{2}}{a^{2}+b^{2}+c^{2}}$, similarly, we have $\frac{b^{4}}{a^{4}+4 b^{4}+c^{4}} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,799 |
Example 4 (1997 USA Mathematical Olympiad) Let $a, b, c > 0$, prove that:
$$\frac{1}{b^{3}+c^{3}+a b c}+\frac{1}{c^{3}+a^{3}+a b c}+\frac{1}{a^{3}+b^{3}+a b c} \leqslant \frac{1}{a b c} .$$ | Proof: $\because b^{3}+c^{3} \geqslant b^{2} c+b c^{2}, \therefore \frac{a b c}{b^{3}+c^{3}+a b c} \leqslant$ $\frac{a b c}{b^{2} c+b c^{2}+a b c}=\frac{a}{a+b+c}$, Similarly, we have $\frac{a b c}{c^{3}+a^{3}+a b c}$ $\leqslant \frac{b}{a+b+c}, \frac{a b c}{a^{3}+b^{3}+a b c} \leqslant \frac{c}{a+b+c}$, Adding the thr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,800 |
Example 5 (2005 Asia Pacific Mathematical Olympiad)
Let
$a, b, c>0, a b c=8$, prove that
$$\begin{array}{l}
\frac{a^{2}}{\sqrt{\left(1+a^{3}\right)\left(1+b^{3}\right)}}+\frac{b^{2}}{\sqrt{\left(1+b^{3}\right)\left(1+c^{3}\right)}}+ \\
\frac{c^{2}}{\sqrt{\left(1+c^{3}\right)\left(1+a^{3}\right)}} \geqslant \frac{4}{3}
... | Proof: By the AM-GM inequality, we have $\sqrt{1+a^{3}}=$ $\sqrt{(1+a)\left(1-a+a^{2}\right)} \leqslant \frac{1+a+1-a+a^{2}}{2}=$ $\frac{a^{2}+2}{2}$, so $\frac{a^{2}}{\sqrt{\left(1+a^{3}\right)\left(1+b^{3}\right)}} \geqslant$ $\frac{4 a^{2}}{\left(a^{2}+2\right)\left(b^{2}+2\right)}$. Similarly, we have $\frac{b^{2}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,801 |
Example 6 (2005 8th Hong Kong Mathematical Olympiad, Problem 3) Let $a, b, c, d>0$ and $a+b+c+d=1$, prove that $6\left(a^{3} + b^{3} + c^{3} + d^{3}\right) \geqslant a^{2} + b^{2} + c^{2} + d^{2} + \frac{1}{8}$. Tangent Line Method | Prove: Let $f(x)=6 x^{3}-x^{2}$, the original inequality is $f(a)+$ $f(b)+f(c)+f(d) \geqslant \frac{1}{8}$, where $a, b, c, d>0$ and $a+$ $b+c+d=1$.
Let $\lambda$ be an undetermined coefficient, such that $f(x)-\frac{1}{32}=6 x^{3}-x^{2}-\frac{1}{32}$ $\geqslant \lambda\left(x-\frac{1}{4}\right)$, this inequality is e... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,802 |
Example 7 (2003 China Western Mathematical Olympiad) $x_{i}>0(i=1,2,3,4,5)$, and $\sum_{i=1}^{n} \frac{1}{1+x_{i}}=1$, prove that $\sum_{i=1}^{5}$
$$\frac{x_{i}}{4+x_{i}^{2}} \leqslant 1$$ | Proof: Let the undetermined constant $\lambda$, such that $\frac{x}{4+x^{2}}-\frac{1}{5} \leqslant$ $\lambda\left(\frac{1}{1+x}-\frac{1}{5}\right) \Leftrightarrow \frac{-(x-4)(x-1)}{x^{2}+4} \leqslant-\lambda \frac{(x-4)}{x+1}$, considering the equality holds, we should have $\frac{(x-4)(x-1)}{x^{2}+4}=-\lambda$ $\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,803 |
Example 8 The 2nd Northern Mathematical Olympiad (2006) Problem 5 Given positive numbers $a, b, c$ satisfying $a+b+c=3$, prove that
$$\begin{array}{l}
\frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2 b^{2}+(c+a)^{2}}+ \\
\frac{c^{2}+9}{2 c^{2}+(a+b)^{2}} \leqslant 5
\end{array}$$ | Proof: Since when $a=b=c=1$, (1) holds with equality, and $\frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}=\frac{5}{3}$, we can set $A$ as a constant to be determined, such that $\frac{a^{2}+9}{2 a^{2}+(3-a)^{2}}-\frac{5}{3} \leqslant A(a-1) \Leftrightarrow$ $\frac{-2(2 a-3)(a-1)}{3\left(a^{2}-2 a+3\right)} \leqslant A(a-1)$. Consid... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,804 |
Example 9 (2003 32nd US Mathematical Olympiad Problem) Let $a, b, c > 0$, prove that $\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(2 b+c+a)^{2}}{2 b^{2}+(c+a)^{2}}$ $+\frac{(2 c+a+b)^{2}}{2 c^{2}+(a+b)^{2}} \leqslant 8$ | Prove: Since the numerator and denominator of each term on the left side of the inequality are homogeneous, we may assume \(a+b+c=3\), then
$$\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}=\frac{(a+3)^{2}}{2 a^{2}+(3-a)^{2}} \leqslant \frac{4}{3} a+\frac{4}{3}$$
In fact, $(*) \Leftrightarrow 4 a^{3}-5 a^{2}-2 a+3 \geqslant 0... | 8 | Inequalities | proof | Yes | Yes | inequalities | false | 737,805 |
Example 10 (1997 Japan Mathematical Olympiad) Let $a, b, c > 0$, prove that $\frac{(b+c-a)^{2}}{(b+c)^{2}+a^{2}}+\frac{(c+a-b)^{2}}{(c+a)^{2}+b^{2}}+$ $\frac{(a+b-c)^{2}}{(a+b)^{2}+c^{2}} \geqslant \frac{3}{5}$ | Proof: Since the numerator and denominator of each term on the left side of the inequality are homogeneous, we can assume $a+b+c=3$, then the original inequality is equivalent to $\frac{(3-2 a)^{2}}{(3-a)^{2}+a^{2}}+\frac{(3-2 b)^{2}}{(3-b)^{2}+b^{2}}+\frac{(3-2 c)^{2}}{(3-c)^{2}+c^{2}} \geqslant \frac{3}{5}$. Let the ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,806 |
4 Try to find a set of positive integer solutions for the equation $x^{2}-51 y^{2}=1$ | $$\begin{array}{l}
\text { Hint: } x^{2}=51 y^{2}+1 \\
=49 y^{2}+14 y+1+2 y^{2}-14 y \\
=\left(7 y+1\right)^{2}+2 y(y-7) \\
\end{array}$$
Let $y=7$. Then $x=50$ | x=50, y=7 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,807 |
6 Prove: There are infinitely many positive integers $n$, such that the average $\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}$ is a perfect square. The first such number is of course 1. Please write down the two positive integers immediately following 1. | $$\begin{array}{l}
\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}=\frac{2 n^{2}+3 n+1}{6} . \\
\text { Let } \frac{1^{2}+2^{2}+\cdots+n^{2}}{n}=m^{2} \text {. Then } \\
2 n^{2}+3 n+1=6 m^{2},
\end{array}$$
i.e. $\square$
$$\left(4 n+3\right)^{2}-3\left(4 n^{2}+1\right)$$
Let $x=4 n+3, y=4 m$, we get the Pell equation
$$x^{2}-3 y... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 737,809 |
Example 3 ( $n$-variable fractional inequality Given $x_{\mathrm{b}}, x_{2}$, $\cdots, x_{n} \in \mathbf{R}_{+}$. Prove:
$$\sum_{i=1}^{n} \frac{x_{i}^{m}}{\sum_{j=1}^{n} x_{j}-x_{i}} \geqslant \frac{A^{m-1}}{n^{m-2}(n-1)}$$
where, $A=x_{1}+x_{2}+\cdots+x_{n}, m \geqslant 2$ | $$\begin{array}{l}
\text { Prove } \sum_{i=1}^{n} \frac{x_{i}^{m}}{\sum_{j=1}^{n} x_{j}-x_{i}}+\frac{A^{m-1}}{\left(n-1 n^{m-2}\right.} \\
=\sum_{i=1}^{n}\left[\frac{x_{i}^{m}}{\sum_{j=1}^{n} x_{j}-x_{i}}+\frac{\left(\frac{A}{n}\right)^{m-2}\left(\sum_{j=1}^{n} x_{j}-x_{i}\right.}{\left(n-1^{2}\right.}\right] \\
\geqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,812 |
Example 1 Suppose the sum of $m$ distinct positive even numbers and $n$ distinct positive odd numbers is 1987, find the maximum value of $3m + 4n$. (2nd CMO Problem) | Let the given $m$ positive even numbers be $a_{1}, a_{2}, \cdots, a_{m}$, and $n$ positive odd numbers be $b_{1}, b_{2}, \cdots, b_{n}$, then
$$\left(a_{1}+a_{2}+\cdots+a_{m}\right)+\left(b_{1}+b_{2}+\cdots+b_{n}\right)=1987$$
Notice that the extremal function is a function of $m$ and $n$, and in the constraint, $m$ a... | 221 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,813 |
Example 2 Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}^{+}, \sum_{i=1}^{n} \frac{1}{x_{i}}=A$ (constant), for a given positive integer $k$, find the maximum value of $\sum \frac{1}{x_{i_{1}}+x_{i_{2}}+\cdots+x_{i_{k}}}$. The summation is over all $k-$ tuples $\left(i_{1}, i_{2}, \cdots, i_{k}\right)$ from $1,2, \cdo... | Analysis and Solution The difficulty of this problem lies in the complexity of the objective function, with the expectation to simplify it using inequalities. Given the structural characteristics of the objective function, the idea is to transform $\frac{1}{x_{i_{1}}+x_{i_{2}}+\cdots+x_{i_{k}}}$ into $\frac{1}{x_{i_{1}... | \frac{A}{k^{2}} \mathrm{C}_{n-1}^{k-1} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,814 |
5ـ Let $x_{1}, x_{2}, \cdots, x_{1990}$ be a permutation of $1,2, \cdots, 1990$, find the maximum value of $F=|\cdots| \mid x_{1}-$ $x_{2}\left|-x_{3}\right|-\cdots\left|-x_{1990}\right|$. (24th All-Russian Mathematical Olympiad Problem) | 5. $F_{2}=|1-2|=1 \leqslant 2, F_{3}=|| 1-2|-3|=2 \leqslant 3, F_{4}=|| \mid 1-$ $2|-3|-4 \mid=4 \leqslant 4$. Generally, conjecture $F_{n} \leqslant n$. We will prove this by mathematical induction. First, note that when $x, y > 0$, $|x-y| \leqslant \max \{x, y\}$. Therefore, $\left|x_{1}-x_{2}\right| \leqslant \max \... | 1989 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,815 |
4. There are 16 students taking an exam, and all questions are multiple-choice with 4 options each. After the exam, it is found that: any two students have at most one question with the same answer. How many questions are there at most? (33rd IMO
| 4. Let there be $n$ questions, and we prove that $n_{\text {max }}=5$. Clearly, $n_{\text {max }}>1$. When $n>1$, for a certain question $A$, if 5 people choose the same answer, then these 5 people's answers to any other question $B$ are all different. However, question $B$ only has 4 options, leading to a contradictio... | 5 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,816 |
5 Let the set $M=\{1,2, \cdots, 10\}$ have $k$ 5-element subsets $A_{1}, A_{2}, \cdots, A_{k}$ that satisfy the condition: any two elements in $M$ appear in at most two subsets $A_{i}$ and $A_{j}(i \neq j)$, find the maximum value of $k$.
| 5. Let the number of times an element $i (i=1,2, \cdots, 10)$ in $M$ appears in $A_{1}, A_{2}, \cdots, A_{k}$ be denoted as $d(i)$.
Considering the whole, an obvious equation is: $d(1)+\cdots+d(10)=\left|A_{1}\right|+\left|A_{2}\right|+\cdots+\left|A_{k}\right|=5+5+\cdots+5=5 k$. To find the range of $k$, we only need... | 8 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,817 |
6 A country has $n$ airports, served by $k$ airlines, with some airports connected by direct flights (direct flights are bidirectional, meaning one can travel from $A$ to $B$ and from $B$ to $A$). If two airports are not directly connected, they can still be reached via connecting flights.
(1) To connect these $n$ airp... | 6. (1) Suppose the $i$-th airline ($i=1,2, \cdots, k$) provides $a_{i}$ direct flights, and let $S=a_{1}+a_{2}+\cdots+a_{k}$. When the $i$-th airline goes bankrupt, the number of direct flights provided by the remaining airlines is $S-a_{i}$. Therefore, $S-a_{i} \geqslant n-1$.
Thus, $\sum_{i=1}^{k}\left(S-a_{i}\right... | 11 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,818 |
Example 1 Given that after 20 gymnasts perform, 9 judges respectively assign them ranks from 1 to 20. It is known that, among the 9 ranks each athlete receives, the maximum and minimum differ by at most 3. Now, the sums of the ranks received by each person are arranged as $c_{1} \leqslant c_{2} \leqslant \cdots \leqsla... | Analysis and Solution First, note that the goal of the solution is $c_{1} \leqslant P$, which is equivalent to the existence of an $i$ such that $c_{i} \leqslant P$. Therefore, estimating the total score of any player can provide an estimate for $c_{1}$.
To minimize the score (rank), it is desirable to have more judge... | 24 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,819 |
Example 2 There are $r$ people participating in a chess tournament, every 2 people play one game, the winner of each game gets 2 points, the loser gets 0 points, and in the case of a draw, each gets 1 point. After the tournament, there is exactly one person who has the fewest wins, and only he has the highest score. Fi... | Analysis and Solution: First, consider the condition: "The number of rounds where exactly one person wins is the least, and only he scores the most," let this person be $A$. To calculate $A$'s score, we can introduce parameters: suppose $A$ wins $n$ rounds and draws $m$ rounds, then $A$'s score is $2n + m$.
Next, cons... | 6 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,820 |
Example 3: There are $n(n \geqslant 5)$ football teams participating in a round-robin tournament. Each pair of teams plays one match, with the winning team getting 3 points, the losing team getting 0 points, and each team getting 1 point in the case of a draw. The team that finishes in the third-to-last place scores fe... | Let $A_{1}, A_{2}, \cdots, A_{n-3}$ be the teams with high scores, $B$ be the team with the third-lowest score, and $C_{1}, C_{2}$ be the teams with the lowest scores. Introduce parameters: Let $B$ win $x$ games, draw $y$ games, and lose $z$ games, then $n=x+y+z+1$. Since $C_{1}$ wins at least $(x+1)$ games, we have $3... | 13 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,821 |
Example 4 There are 1000 certificates numbered $000,001, \cdots, 999$ and 100 boxes numbered 00, $01, \cdots, 99$. If the number of a box can be obtained by removing one digit from the number of a certificate, then the certificate can be placed in that box. If $k$ boxes can be chosen to accommodate all certificates, fi... | Analysis and Solution: Find a sufficient condition to ensure that the selected boxes can accommodate all the documents.
Consider all documents with numbers $a$, $b$, and $c$ (where $a$, $b$, and $c$ may be the same) as their codes. To accommodate these documents, there must be a box with at least two of these numbers (... | 50 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,822 |
Example 5 Find the smallest natural number $n$ such that: when any 5 vertices of a regular $n$-gon $S$ are colored red, there is always a line of symmetry $L$ of $S$ such that the reflection of each red point across $L$ is not a red point. | Let the regular $n$-sided polygon be $A_{1}, A_{2}, \cdots, A_{n}$, which has $n$ axes of symmetry.
Let the axis of symmetry passing through the midpoint of edge $A_{1} A_{n}$ be denoted as $L_{1}$, the axis of symmetry passing through $A_{1}$ as $L_{2}$, the axis of symmetry passing through the midpoint of edge $A_{1}... | 14 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,823 |
Example 6 For an integer $n \geqslant 4$, find the smallest integer $f(n)$, such that for any positive integer $m$, in any $f(n)$-element subset of the set $\{m, m+1, \cdots, m+n-1\}$, there are at least 3 pairwise coprime elements. (2004 National High School Mathematics Competition Problem) | Let's solve the problem: When $n \geqslant 4$, for the set $M=\{m, m+1, m+2, \cdots, m+n-1\}$, if $m$ is odd, then $m, m+1, m+2$ are pairwise coprime; if $m$ is even, then $m+1, m+2, m+3$ are pairwise coprime. Thus, every $n$-element subset of $M$ contains at least 3 pairwise coprime numbers, so $f(n)$ exists and $f(n)... | f(n)=\left[\frac{n+1}{2}\right]+\left[\frac{n+1}{3}\right]-\left[\frac{n+1}{6}\right]+1 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,824 |
6 Let $0<p \leqslant a_{i} \leqslant q, b_{i}$ be a permutation of $a_{i}$ $(1 \leqslant i \leqslant n)$, find the extremum of $F=\sum_{i=1}^{n} \frac{a_{i}}{b_{i}}$. (Hungarian Mathematical Olympiad Problem) | 6. Let's assume $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$. Since the function is continuous on a closed interval, $F$ has a maximum and minimum value. By the rearrangement inequality, we have $F=\sum_{i=1}^{n} \frac{a_{i}}{b_{i}} \geqslant \sum_{i=1}^{n} \frac{a_{i}}{a_{i}}=\sum_{i=1}^{n} 1=n$, with equa... | n+\left[\frac{n}{2}\right]\left(\sqrt{\frac{p}{q}}-\sqrt{\frac{q}{p}}\right)^{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,826 |
1 In an exam, there are 30 multiple-choice questions. Correct answers earn 5 points each, incorrect answers earn 0 points, and unanswered questions earn 1 point each. If person A scores more than 80 points, and tells B the score, B can deduce how many questions A answered correctly. If A's score is slightly lower but s... | 1. Jia's score $S$ is determined by his answer situation, so we can introduce parameters: Let the number of questions Jia answered correctly, unanswered, and incorrectly be $x$, $y$, and $z$, respectively, with $x+y+z=30$. Then Jia's score $S=5x+y+0z=5x+y$. The phrase "if Jia's score is lower but still greater than 80,... | 119 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,827 |
2 A class has 47 students, and the classroom used has 6 rows, with 8 seats in each row. The seat located in the $i$th row and the $j$th column is denoted as $(i, j)$. For the new semester, the seats are to be rearranged. If a student's original seat is $(i, j)$ and the new seat is $(m, n)$, then the student's movement ... | 2. The value $i+j$ is called the characteristic value of the cell $(i, j)$. The number of positions a student can move to is the difference between the characteristic values of the cells before and after the move. Let the sum of the characteristic values of all cells be $M$. Introduce parameters: Let the initial empty ... | 12 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,828 |
3 It is known that there are 12 theater troupes participating in a 7-day performance, and it is required that each troupe can see the performances of all other troupes, with only those not performing on a given day being able to watch from the audience. How many performances are needed at a minimum? (1994 China Nationa... | 3. Let $A=\{1,2, \cdots, 7\}$ be the set of performance dates, and $A_{i}$ be the set of performance dates for the $i$-th troupe $(i=1,2, \cdots, 12)$. It is clear that $A_{1}, A_{2}, \cdots, A_{12}$ do not contain each other. Otherwise, for $i \neq j$, if $A_{i}$ is contained in $A_{j}$, then the $j$-th troupe cannot ... | 22 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,829 |
4. For each positive integer $n$, let $s(n)$ denote the greatest integer such that for any positive integer $k \leqslant s(n)$, $n^2$ can be expressed as the sum of $k$ positive integer squares. (1) Prove that $s(n) \leqslant n^2 - 14$ for $n \geqslant 4$. (2) Find an $n$ such that $s(n) = n^2 - 14$. (3) Prove that the... | 4. (1) Suppose $n^{2}=a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2}$, where $k=n^{2}-13$. Without loss of generality, assume $a_{1} \leqslant a_{2} \leqslant a_{3} \leqslant \cdots \leqslant a_{k}$, then $a_{k}^{2}=n^{2}-\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{k-1}^{2}\right) \leqslant n^{2}-(k-1)=n^{2}-\left(n^{2}-14\right)=14$. T... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 737,830 |
Example 1 In a certain parliament, there are 30 deputies, and any two deputies are either political enemies or friends. Moreover, each deputy has exactly 6 political enemies. For a committee of 3 deputies, if any two of the three are friends or any two are political enemies, it is called a strange committee. How many s... | Solve: This problem has a clear graph theory flavor: either friends or political enemies - connecting edges or not (coloring red or blue); each senator has 6 political enemies - each point leads to 6 red edges; the peculiar committee - monochromatic triangles.
Consider a complete graph with 30 points. Each point repre... | 1990 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,832 |
Example 2: There are 18 points on a plane, no three of which are collinear. Each pair of points is connected by a line segment, which is colored either red or blue, with each line segment being only one color. It is known that a certain point $A$ has an odd number of red line segments emanating from it, and the number ... | Remove all blue edges from the graph, obtaining a simple graph $G$. By the given conditions, $d(A)$ is odd, and the degrees of other non-$A$ points belong to $\{0,1,2, \cdots, 17\}$. It is easy to see that points with degree 0 and 17 cannot coexist. Therefore, the degrees of the points are either $0,1,2, \cdots, 16$ or... | 204 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,833 |
Example 3 Let $S=\{1,2, \cdots, 15\}$. From $S$, take $n$ subsets $A_{1}, A_{2}, \cdots, A_{n}$, satisfying the following conditions:
(1) $\left|A_{i}\right|=7(i=1,2, \cdots, n)$;
(2) $\left|A_{i} \cap A_{j}\right| \leqslant 3(1 \leqslant i<j \leqslant n)$;
(3) For any 3-element subset $M$ of $S$, there exists some $A_... | Analysis and Solution: From condition (1) $\left|A_{i}\right|=7(i=1,2, \cdots, n)$, we think of calculating the total number of times each element appears in all subsets $S_{1}$.
On the one hand, starting from each subset, we have $S_{1}=7 n$. On the other hand, starting from each element, let the number of times $i(i... | 15 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,834 |
There are 8 singers participating in the arts festival, and today we need to arrange $m$ performances, each time with 4 of them performing on stage. It is required that any two of the 8 singers perform together the same number of times. Please design a scheme so that the number of performances $m$ is minimized (Problem... | Analysis and Solution: First, consider the condition: "arrange $m$ performances, each time 4 of them perform on stage", which is equivalent to: there are $m$ subsets, each set has 4 elements. Then, consider the condition: "the number of times any two perform together is the same", which is equivalent to: each 2-element... | null | Geometry | math-word-problem | Yes | Yes | inequalities | false | 737,835 |
Example 5: On the ground, there are 10 birds pecking, and among any 5 birds, at least 4 birds are on the same circumference. What is the maximum number of birds on the circumference that has the most birds? (6th China Mathematical Olympiad) | Analysis and Solution: Use points to represent birds. Let the circle with the most points have $r$ points. Clearly, $r \geqslant 4$. Can $r=4$? Let's first explore the case where $r=4$.
If $r=4$, i.e., each circle has at most 4 points, but we are considering circles that pass through at least 4 points, so each "4-poin... | 9 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,836 |
Example 1 Let $x, y, z$ be non-negative real numbers, $x+y+z=1$, find the maximum and minimum values of $F=2 x^{2}+y+3 z^{2}$.
| First, using substitution and elimination, we have $y=1-x-z$, so
$$\begin{aligned}
F & =2 x^{2}+1-x-z+3 z^{2} \\
& =2\left(x-\frac{1}{4}\right)^{2}+3\left(z-\frac{1}{6}\right)^{2}+\frac{19}{24} \\
& \geqslant \frac{19}{24}
\end{aligned}$$
Also, $F\left(\frac{1}{4}, \frac{7}{12}, \frac{1}{6}\right)=\frac{19}{24}$, so t... | \frac{19}{24}, 3 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,837 |
Example 6 There are 16 students taking an exam, all questions are multiple-choice, each with 4 options. After the exam, it was found that: any two students have at most one question with the same answer. How many questions are there at most? (33rd
| Analysis and Solution: Let there be $n$ questions. We prove that $n_{\text {max }}=5$.
For each question, the answers of 16 students form a sequence of length 16, containing only $1, 2, 3, 4$ (answer codes). Arrange the sequences corresponding to $n$ questions into an $n \times 16$ number table.
Notice the condition "... | 5 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,838 |
Example 7 $n$ people greet each other by phone during a holiday, it is known that each person has called at most 3 of their friends; any 2 people have had at most 1 phone call; and among any 3 people, at least 2 of them, one has called the other. Find the maximum value of $n$. (Original problem) | Solve the original problem: Use $n$ points to represent $n$ people. If person $A$ calls person $B$'s home phone, then draw a directed edge from $A$ to $B$, resulting in a simple directed graph $G$.
On one hand, $\bar{G}$ contains no triangles, and by the lemma, $|| \bar{G} || \leqslant \left[\frac{n^{2}}{4}\right]$, s... | 14 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,839 |
Example 8 There are three people, $A$, $B$, and $C$, playing table tennis. When two of them play, the third one acts as the referee. In this match, the loser will be the referee in the next match, and the other two will continue to play. After several matches, it is known that $A$ has played a total of $a$ matches, and... | Let $C$ have played $c$ games, then the total number of player appearances is $a+b+c$, but each game produces 2 player appearances, so the total number of games played is $\frac{a+b+c}{2}$.
Therefore, the number of games where $C$ was the referee is $\frac{a+b+c}{2}-c=\frac{a+b-c}{2}$.
Since if $C$ is the referee in a... | c_{\min }=\left\{\begin{array}{ll}{\left[\frac{a+b}{3}\right]} & (a+b \equiv 0,2(\bmod 3) ; \\ {\left[\frac{a+b}{3}\right]+1} & (a+b \equiv 1(\bmod 3)\end{array}\right.} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,840 |
Example 9 Let $|X|=56$, for any 15 subsets of $X$, if the union of any 7 of them contains at least $n$ elements, then among these 15 subsets, there must exist 3 subsets whose intersection is non-empty, find the minimum value of $n$. (2006 China Mathematical Olympiad Problem) | Solve $n_{\min }=41$.
First, prove that $n=41$ meets the conditions. Use proof by contradiction: Assume there exist 15 subsets of $X$ such that the union of any 7 of them contains at least 41 elements, and the intersection of any 3 of them is empty. Then each element belongs to at most 2 subsets. Without loss of genera... | 41 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,841 |
1650 students are arranged in 22 rows and 75 columns. It is known that among any two columns in the same row, the number of pairs of students of the same gender does not exceed 11. Prove: the number of boys does not exceed 928. (2003
| 1. Let the number of boys in the $i$-th row be $a_{i}$, then the number of girls is $75-a_{i}$. According to the problem, we have $\sum_{i=1}^{22}\left(\mathrm{C}_{a_{i}}^{2}+\right.$ $\left.\mathrm{C}_{75-a_{i}}^{2}\right) \leqslant 11 \times \mathrm{C}_{75}^{2}$. This is because, for any two columns, the number of pa... | 928 | Combinatorics | proof | Yes | Yes | inequalities | false | 737,842 |
2 A meeting has $12 k$ people attending, each person has greeted exactly $3 k+6$ people. For any two people, the number of people they have greeted is the same. How many people attended this meeting? | 2. Represent people with dots, and for two people who have greeted each other, connect the corresponding two dots to form a simple graph. Calculate the number of angles $S$ in this graph. On one hand, for $G$, each point leads to $3k+6$ edges, resulting in $\mathrm{C}_{3k+6}^{2}$ angles. With $12k$ points, there are $1... | 36 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,843 |
3 There are $n$ players participating in a competition lasting $k$ days, where the scores of the $n$ players on any given day are exactly a permutation of $1,2,3, \cdots, n$. If at the end of the $k$th day, each player's total score is 26. Find all possible values of $(n, k)$.
| 3. From the condition "each player's total score is 26", we think of calculating the sum of scores $S$ of $n$ players after $k$ days. On one hand, the score each day is $1+2+3+\cdots+n$, so $S=k(1+2+\cdots+n)$. On the other hand, each player scores 26 points, thus $S=26n$. Therefore, $k(n+1)=52$. So $(n, k)=$ $(51,1) 、... | (25,2), (12,4), (3,13) | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,844 |
4 A class has 30 students, all of different ages, and each student has the same number of friends within the class. For a student $A$, if $A$ is older than more than half (not including half) of $A$'s friends, then $A$ is called elderly. What is the maximum number of elderly students? (20th All-Russian Mathematical Oly... | 4. Suppose there are $t$ older students, each of whom has $k$ friends. For the convenience of describing the problem, use 30 points to represent 30 students. For any two points $A$ and $B$, if $A$ and $B$ are friends, and $A$ is older than $B$, then draw a directed edge pointing to $B$, resulting in a tournament graph.... | 25 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,845 |
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