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Question 8 Let $a, b, c \in \mathbf{R}_{+}$, and $a^{2}+b^{2}+c^{2}+a b c=4$, prove: $\frac{a b}{a^{2}+4}+\frac{b c}{b^{2}+4}+\frac{c a}{c^{2}+4} \leqslant \frac{3}{5}$. | Prove: In the proof of problem 7, we obtained
$$\frac{t}{t^{2}+4} \leqslant \frac{3 t}{25}+\frac{2}{25}(0<t<2)$$
Taking \( t=a \), we get \(\frac{a}{a^{2}+4} \leqslant \frac{3 a}{25}+\frac{2}{25}\),
then \(\frac{a b}{a^{2}+4} \leqslant \frac{3 a b}{25}+\frac{2 b}{25}\).
Similarly, we get \(\frac{b c}{b^{2}+4} \leqslan... | \frac{3}{5} | Inequalities | proof | Yes | Yes | inequalities | false | 736,300 |
Question 9 Let $a, b, c \in \mathbf{R}_{+}$, and $a^{2}+b^{2}+c^{2}+a b c=4$, prove: $a+b+c>2$.
| $$\begin{array}{l}
\text { From the given conditions, it is easy to know } 0 < a, b, c < 1 \\
\text { and } a+b, b+c, c+a > 1 \text {. }
\end{array}$$
Thus, we have $(a+b-1)+(b+c-1)+(c+a-1)>0$,
which means $2(a+b+c)-3>0$,
and therefore $2(a+b+c)>3$,
which implies $(a+b+c)>\frac{3}{2}$.
Since $a, b, c < 1$, we have... | a+b+c>2 | Inequalities | proof | Yes | Yes | inequalities | false | 736,301 |
Question 1 Given that $x, y, z$ are positive real numbers, prove:
$$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \frac{3}{2}$$ | This problem was once a question in the 1963 Moscow Competition, and it has many proof methods. Here, we use local substitution and combine it with determinant knowledge to provide a clever transformation.
If we let \( a = \frac{x}{y+z}, b = \frac{y}{z+x}, c = \frac{z}{x+y} \) \(\left(a, b, c \in \mathbf{R}^{+}\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,302 |
Question 2 Given that $a$, $b$, $c$ are positive real numbers, $ab + bc + ca + 2abc = 1$, prove:
$$a + b + c \geqslant \frac{3}{2}$$ | $$\begin{array}{l}
\text { Proof: Applying the AM-GM inequality, we get } \\
a b+b c+c a \leqslant \frac{1}{3}(a+b+c)^{2}, \\
a b c \leqslant\left(\frac{a+b+c}{3}\right)^{3},
\end{array}$$
Combining these inequalities with the given equation, we can derive the inequality
$$\frac{1}{3}(a+b+c)^{2}+2\left(\frac{a+b+c}{3}... | a + b + c \geqslant \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,303 |
Question 3 Given that $a, b, c$ are positive real numbers, $ab + bc + ca + 2abc = 1$, prove:
$$ab + bc + ca \geqslant \frac{3}{4}$$ | Proof: By the 3-variable mean inequality, we get
$$(a b c)^{2}=a b \cdot b c \cdot c a \leqslant\left(\frac{a b+b c+c a}{3}\right)^{3}$$
which implies \(a b c \leqslant\left(\frac{a b+b c+c a}{3}\right)^{\frac{3}{2}}\).
Combining this with the given equality, we can derive the inequality \(a b+b c+c a+2\left(\frac{a b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,304 |
Question 4 Given that $a$, $b$, $c$ are positive real numbers, $a b + b c + c a + 2 a b c = 1$, prove: the pigeonhole principle, I don't understand the underlying concept
$$a + b + c \geqslant 2(a b + b c + c a)$$ | Proof: Since among the three numbers $a$, $b$, and $c$, there must be two on the same side of $\frac{1}{2}$, we can assume without loss of generality that $b$ and $c$ are on the same side of $\frac{1}{2}$. Therefore, we have
$$\left(b-\frac{1}{2}\right)\left(c-\frac{1}{2}\right) \geqslant 0$$
which implies $(2 b-1)(2 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,305 |
Question 1 In an acute $\triangle A B C$, we have
$$\sum \frac{1}{\sin 2 A} \geqslant \sum \frac{1}{\sin A}$$
Equality holds if and only if $\triangle A B C$ is an equilateral triangle.
After thinking and exploring, the author has derived a stronger conclusion than inequality (1):
Question 2 In an acute $\triangle A B... | $$\begin{array}{l}
\text { Proof: From the binary mean inequality } 4 x y \leqslant(x+y)^{2} \text {, we get } \\
\sin 2 A \sin 2 B=4(\sin A \cos B)(\cos A \sin B) \\
\leqslant(\sin A \cos B+\cos A \sin B)^{2} \\
=\sin ^{2}(A+B) \\
=\sin ^{2} C,
\end{array}$$
Therefore, $\frac{1}{\sin 2 A \sin 2 B} \geqslant \frac{1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,308 |
$$\begin{array}{l}
\cos \alpha+\cos \left(120^{\circ}-\alpha\right)+\cos \left(120^{\circ}+\alpha\right)=0 \\
\sin \alpha-\sin \left(120^{\circ}-\alpha\right)+\sin \left(120^{\circ}+\alpha\right)=0
\end{array}$$
where $\alpha$ is any angle. | $$\begin{array}{l}
P=\cos \alpha+\cos \left(120^{\circ}-\alpha\right)+\cos \left(120^{\circ}+\alpha\right) \\
Q=\sin \alpha-\sin \left(120^{\circ}-\alpha\right)+\sin \left(120^{\circ}+\alpha\right)
\end{array}$$
Let $z_{1}=\cos \alpha+i \sin \alpha, z_{2}=\cos 120^{\circ}+i \sin 120^{\circ}$, then it is not difficult ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,311 |
$\begin{array}{l} \text { Given } a>0, b>0, a+b=1, \\ \sqrt{2}<\sqrt{a+\frac{1}{2}}+\sqrt{b+\frac{1}{2}} \leqslant 2 .\end{array}$ | $$\begin{array}{l}
\text { Given that } 0<\frac{a}{\sqrt{1+\frac{1}{2}}+\sqrt{\frac{1}{2}}}=\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\right)_{a}, \\
\text { i.e., } \sqrt{a+\frac{1}{2}}>\sqrt{\frac{1}{2}}+\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\right)_{a}, \\
\text { similarly, } \sqrt{b+\frac{1}{2}}>\sqrt{\frac{... | \sqrt{a+\frac{1}{2}}+\sqrt{b+\frac{1}{2}}>\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,312 |
Let $x, y, z \in \mathbf{R}_{+}$, then we have
$$\sqrt{\frac{x}{y+z}}+\sqrt{\frac{y}{z+x}}+\sqrt{\frac{z}{x+y}}>2 .$$
Interestingly, Mr. Liu Baoqian conjectured a "sister" inequality to (1) using Model VII in [1]:
Let $x, y, z \in \mathbf{R}_{+}$, then we have
$$\sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z... | Proof: Without loss of generality, assume $x+y+z=1$, otherwise make the substitution $(x, y, z) \rightarrow\left(\frac{x}{x+y+z}, \frac{y}{x+y+z}, \frac{z}{x+y+z}\right)$.
Then $2 \sqrt{x} \cdot \sqrt{1-x} \leqslant 1 \Leftrightarrow \sqrt{\frac{x}{y+z}}=$ $\sqrt{\frac{x}{1-x}} \geqslant 2 x$, equality holds if and on... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,313 |
Promotion 1 Let $x_{i} \in \mathbf{R}_{+}(i=1,2, \cdots n)$, then
$$\sum_{j=1}^{n} \sqrt{\frac{x_{j}}{\sum_{i=1}^{n} x_{i}-x_{j}}}>2$$
Promotion 1 can be weakened to:
Corollary 1 Let $x_{i} \in \mathbf{R}_{+}(i=1,2, \cdots n)$, then
$$\sqrt{\frac{x_{1}}{x_{2}+x_{3}}}+\sqrt{\frac{x_{2}}{x_{3}+x_{4}}}+\cdots+\sqrt{\frac... | Prove that, without loss of generality, let $x+y+z=1$, then
$$\begin{array}{l}
(n-1) \cdot x^{n-1} \cdot(1-x) \leqslant\left(\frac{n-1}{n}\right)^{n} \\
\Leftrightarrow \sqrt[n]{\frac{x}{y+z}}=\sqrt[n]{\frac{x}{1-x}} \geqslant \frac{n \cdot \sqrt[n]{n-1}}{n-1} \cdot x
\end{array}$$
Equality holds if and only if $x=\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,314 |
Question: Let the real constant $a$ be such that the inequality $\frac{1}{1+\sqrt{x}} \geqslant a \cdot \sqrt{\frac{x}{x-1}}$ has non-zero real solutions for $x$. Prove that $a < \frac{1}{3}$.
This problem was designed by Mr. Wu Weizhao. In the background of the problem, he mentions that he has not yet been able to de... | Suppose there is a non-zero real solution $x=x_{0}$, then it is easy to know that: $x_{0}$ $>1$, and we have
$$a \leqslant \frac{1}{1+\sqrt{x_{0}}} \cdot \sqrt{\frac{x_{0}-1}{x_{0}}} \leqslant \max _{x>1}\left\{\frac{1}{1+\sqrt{x}} \cdot \sqrt{\frac{x-1}{x}}\right\} .$$
Let $f(x)=\frac{1}{1+\sqrt{x}} \cdot \sqrt{\frac... | \frac{(\sqrt{5}-2) \sqrt{2 \sqrt{5}+2}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,316 |
Lemma If $f(x)=\frac{1}{1+x}, x \in(0,1), n \in N^{*}, n \geq k+1, k \geq 2$, then
$$f(x) \leq \frac{k n^{K+1}}{\left(1+n^{K}\right)^{K}} x+\frac{n^{K}\left(n^{K}+k+1\right)}{\left(1+n^{K}\right)^{2}}$$
Equality holds if and only if $x=\frac{1}{n}$. | Lemma Proof:
Inequality (4) is equivalent to
$$(1+n k)^{2} \leq -k^{R+1} x\left(1+x^{k}\right)+n^{k}\left(n^{k}+k+1\right)\left(1+x^{k}\right)$$
which is $k n^{k+1} x^{k+1}-n^{k}\left(n^{k}+k+1\right) x^{k}+k n^{k+1} x-(k-1) n^{k}+1 \leq 0$.
$$\text { Let } g(x)=k n^{k+1} x^{k+1}-n^{k}\left(n^{k}+k+1\right) x^{k}+k n^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,321 |
Example 2 Find the minimum value of $y=\sqrt{x^{2}+1}+\sqrt{(4-x)^{2}+4}$. | Solution: Construct right triangles $\triangle P A C$ and $\triangle P B D$ such that $A C=1, B D=$ $2, P C=x, P D=4-x$, and transform the problem of finding the minimum value to: finding a point $P$ on $l$ such that the sum $P A+P B$ is minimized.
Take the symmetric point $A^{\prime}$ of $A$ with respect to $l$, then... | 5 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,324 |
Theorem Let $a, b, c$ be positive real numbers, and $\lambda, \mu, v$ be non-negative real numbers, not all zero, then
$$\sum \frac{a^{2}}{\lambda a+\mu b+v c} \geqslant \frac{a+b+c}{\lambda+\mu+v}$$
where $\sum$ denotes the cyclic sum over $a, b, c$.
Strengthening If $a, b, c \in \mathrm{R}^{+}$, then
$$\sum \frac{a^... | Prove that since $x^{2}+y^{2}-2 x y=(x-y)^{2}$, therefore
$$\frac{x^{2}}{y}=2 x-y+\frac{(x-y)^{2}}{y} .$$
Thus,
$$\frac{(\lambda+\mu+v)^{2} a^{2}}{\lambda a+\mu b+v c}=2(\lambda+\mu+v) a-(\lambda a+\mu b+v c)+\frac{[(\lambda+\mu+v) a-(\lambda a+\mu b+v c)]^{2}}{\lambda a+\mu b+v c}$$
That is,
$$\frac{a^{2}}{\lambda a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,326 |
Conjecture 2 Let $a_{i}(i=1,2, \cdots, n) \in \mathrm{R}^{+}, n \in \mathrm{N}_{+}$, and $n \geqslant 3, a_{1} a_{2} \cdots a_{n}=1, a=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}$. Prove:
$$\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right) \leqslant(1+a)^{n}-a^{n}+1 .$$ | Conjecture 2 Proof: From the problem, we have
$$\frac{s_{1}}{C_{n}^{1}}=a, s_{n}=\prod_{k=1}^{n} a_{k}=1 .$$
By the lemma and the binomial theorem, we get
$$\begin{aligned}
(1 & \left.+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right) \\
& =1+s_{1}+s_{2}+\cdots+s_{n} \\
\leqslant & 1+n a+C_{n}^{2} a^{2}+C_{n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,327 |
Example 2 (Question 14 of the 2008 National College Entrance Examination, Liberal Arts, First Paper) Given that the focus of the parabola $y=a x^{2}-1$ is the origin, then the area of the triangle formed by the three intersection points of the parabola with the two coordinate axes is $\qquad$ . | Solution: Since the parabola equation can be transformed into $x^{2}=\frac{1}{a}(y+$ $1)$, and it is known that its graph intersects the $y$-axis at point $C(0,-1)$, and the focus is the origin, by the geometric properties of the parabola, we know $\frac{p}{2}$ $=1 \Rightarrow p=2$, so $\frac{1}{a}=2 p=4 \Rightarrow a=... | 2 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,331 |
Theorem Let $a, b, c$ be positive real numbers, and $\lambda, \mu, \nu$ be non-negative real numbers, not all zero. Then we have
$$\sum \frac{a^{2}}{\lambda a+\mu b+\nu c} \geqslant \frac{a+b+c}{\lambda+\mu+\nu}(1) .$$
where $\sum$ denotes the cyclic sum over $a, b, c$.
After exploration, the author has derived a stre... | Proof: Since $x^{2}+y^{2}-2xy=(x-y)^{2}$, we have $\frac{x^{2}}{y}=2x-y+\frac{(x-y)^{2}}{y}$. Therefore,
$$\begin{array}{l}
\frac{(\lambda+\mu+\nu)^{2}a^{2}}{\lambda a+\mu b+\nu c}=2(\lambda+\mu+\nu)a-(\lambda a+\mu b \\
+\nu c)+\frac{[(\lambda+\mu+\nu)a-(\lambda a+\mu b+\nu c)]^{2}}{\lambda a+\mu b+\nu c}, \\
\frac{a^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,332 |
Given $\cos \left(\frac{\pi}{4}+x\right)=\frac{3}{5}, \frac{17 \pi}{12}<x<\frac{7 \pi}{4}$, find $\frac{\sin 2 x+2 \sin ^{2} x}{1-\tan x}$. | $$\begin{array}{l}
\text { Method 1: } \because \cos \left(\frac{\pi}{4}+x\right)=\frac{3}{5}, \quad \therefore \cos x \\
-\sin x=\frac{3}{5} \sqrt{2}, \quad 1-2 \sin x \cos x=\frac{18}{25}, \quad(\sin x \\
+\cos x)^{2}=1+2 \sin x \cos x=\frac{32}{25} \text {, and } \frac{17 \pi}{12}<x \\
<\frac{7 \pi}{4} \text {, so }... | \frac{2 \times \left(-\frac{7 \sqrt{2}}{10}\right) \times \left(-\frac{\sqrt{2}}{10}\right) + 2 \times \left(-\frac{7 \sqrt{2}}{10}\right)^2}{1 - 7} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,333 |
Suppose $x, y, z \in R^{+}$, then
$$\sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}} \leqslant \frac{3 \sqrt{2}}{2} .$$
In article [2], Mr. Wu Shanhe provided a simple proof of (1) using the binary mean inequality. In article [3], Mr. Shu Jingen also proved inequality (1) using the mean inequality and de... | Let $a=\frac{y}{x}, b=\frac{z}{y}, c=\frac{x}{z}$, then $a, b, c \in R^{+}, a b c=1$, so the inequality (2) is equivalent to
$$\frac{1}{\sqrt{a+2}}+\frac{1}{\sqrt{b+2}}+\frac{1}{\sqrt{c}+2} \leqslant \sqrt{3}$$
By the Cauchy-Schwarz inequality, we have
$$\frac{1}{\sqrt{a+2}}+\frac{1}{\sqrt{b+2}}+\frac{1}{\sqrt{c+2}} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,334 |
Example 3 In $\triangle A B C$, $D$ is a point on side $B C$, $B D=$ $\frac{1}{2} D C, \angle A D B=120^{\circ}, A D=2$, if the area of $\triangle A D C$ is $3-\sqrt{3}$, then $\angle B A C=$ $\qquad$ | Solve: Since $\angle A D B=120^{\circ}$, and $B D=\frac{1}{2} D C$, it is easy to know that $S_{\triangle A B D}=\frac{1}{2} S_{\triangle A D C}=\frac{1}{2}(3-$
Figure 3 $\sqrt{3})$. Also, $S_{\triangle A B D}=\frac{1}{2} B D \cdot A D \cdot \sin 120^{\circ}$, so we have $B D \cdot \frac{\sqrt{3}}{2}=\frac{1}{2}(3-\sq... | 60^{\circ} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,335 |
If $x, y, z$ are positive numbers, and $x y+y z+z x=$ 1, then we have
$$8 x^{2} y^{2} z^{2} \geqslant\left(1-x^{2}\right)\left(1-y^{2}\right)\left(1-z^{2}\right)(*)$$ | Proof: From the given conditions, if two of $x, y, z$ are not less than 1, then we have $1 = xy + yz + zx > 1$, which leads to a contradiction. Therefore, at most one of $x, y, z$ is not less than 1.
(1) If one of $x, y, z$ is not less than 1, in this case, the left side of inequality $(*)$ is $>0$, while the right si... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,336 |
If $a_{1}, a_{2}, a_{3}, a_{4} \in \mathbf{R}^{+}$, prove:
$$\begin{array}{l}
\frac{a_{1}^{3}}{a_{2}+a_{3}+a_{4}}+\frac{a_{2}^{3}}{a_{1}+a_{3}+a_{4}} \\
+\frac{a_{3}^{3}}{a_{1}+a_{2}+a_{4}}+\frac{a_{4}^{3}}{a_{1}+a_{2}+a_{3}} \\
\geqslant \frac{\left(a_{1}+a_{2}+a_{3}+a_{4}\right)^{2}}{12}
\end{array}$$
This article w... | Proof: $\because \frac{a^{2}}{b} \geqslant 2 a-b\left(a 、 b \in \mathbf{R}^{+}\right)$,
$$\begin{array}{c}
o \\
\therefore \frac{\left(3 a_{1}\right)^{2}}{a_{2}+a_{3}+a_{4}} \geqslant 2\left(3 a_{1}\right)-\left(a_{2}+a_{3}+a_{4}\right),
\end{array}$$
i.e., $\frac{a_{1}^{3}}{a_{2}+a_{3}+a_{4}} \geqslant \frac{2}{3} a_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,337 |
$$\begin{array}{l}
\text { Promotion } 1 \text { If } a_{t} \in \mathbf{R}^{+}(i=1,2,3, \cdots, n), \\
s=\sum_{i=1}^{n} a_{i}, \text { and } 2 \leqslant n \in \mathbf{N}, \text { prove: } \\
\sum_{i=1}^{n} \frac{a_{i}^{3}}{s-a_{i}} \geqslant \frac{1}{n-1} \sum_{i=1}^{n} a_{i}^{2} .
\end{array}$$ | Proof: By Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(s-a_{i}\right)^{2} \leqslant(n-1)\left[\sum_{j=1}^{n} a_{j}^{2}-a_{i}^{2}\right] \\
\because \frac{a_{i}^{3}}{s-a_{i}}+\frac{a_{i}^{3}}{s-a_{i}}+\frac{\left(s-a_{i}\right)^{2}}{(n-1)^{3}} \\
\quad \geqslant 3 \sqrt[3]{\left(\frac{a_{i}^{3}}{s-a_{i}}\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,338 |
Promotion 2 If $a_{i} \in \mathbf{R}^{+}(i=1,2,3, \cdots, n)$, $2 \leqslant n \in \mathbf{N}$, and $s=\sum_{i=1}^{n} a_{i}, m \in \mathbf{N}$, prove:
$$\sum_{i=1}^{n} \frac{a_{t}^{m}}{s-a_{i}} \geqslant \frac{1}{n-1} \sum_{i=1}^{n} a_{t}^{m-1}$$ | $$\begin{array}{l}
\left(s-a_{i}\right)^{2} \leqslant(n-1)\left[\sum_{j=1}^{n} a_{j}^{2}-a_{i}^{2}\right] \Rightarrow \\
\left(s-a_{i}\right)^{m-1} \leqslant(n-1)^{\frac{m-1}{2}}\left(\sum_{j=1}^{n} a_{j}^{2}-a_{i}^{2}\right)^{\frac{m-1}{2}} \\
\because \underbrace{\frac{a_{i}^{m}}{s-a_{i}}+\cdots+\frac{a_{i}^{m}}{s-a_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,339 |
Lemma If $f(x)=\frac{1}{\left(1+x^{2}\right)^{2}}, x \in(0,1)$, then we have
$$f(x) \leqslant-\frac{4096}{4913} x+\frac{5376}{4913}$$ | Prove that obviously, inequality (3) is equivalent to
$$4913 \leqslant(-4096 x+5376)\left(1+x^{2}\right)^{2} \text {, }$$
Transforming, we get $4096 x^{5}-5376 x^{4}+8192 x^{3}-10752 x^{2}$
$$\begin{array}{l}
+4096 x-463 \leqslant 0 . \\
\text { And } 4096 x^{5}-5376 x^{4}+8192 x^{3}-10752 x^{2}+4096 x \\
-463=\left(4... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,340 |
Given $x>0, y>0$, and $x+y=1$, prove that
$$(\sqrt{x}+\sqrt{y})\left(\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}\right) \leqslant \frac{4}{\sqrt{3}} .$$ | Proof: From the given,
(1) $\Leftrightarrow(1+2 \sqrt{x y})\left(\frac{3}{2+x y}+\frac{2}{\sqrt{2+x y}}\right) \leqslant \frac{16}{3}$
$$\Leftrightarrow(1+u)\left(\frac{3}{8+u^{2}}+\frac{1}{\sqrt{8+u^{2}}}\right) \leqslant \frac{4}{3}$$
(where $u=2 \sqrt{x y}$)
$$\begin{array}{l}
\Leftrightarrow 3(1+u) \sqrt{8+u^{2}} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,341 |
$$\begin{aligned}
& \text { Given } a>0, b>0 \text {, prove that } \\
& \sqrt{\frac{a}{2 a+b}}+\sqrt{\frac{b}{a+2 b}} \leqslant \sqrt{\frac{a}{a+2 b}}+\sqrt{\frac{b}{2 a+b}} \\
\leqslant & \frac{2}{\sqrt{3}} .
\end{aligned}$$ | Proof: First, prove the left inequality, eliminate the denominator
$$\begin{array}{l}
\Leftrightarrow \sqrt{a(a+2 b)}+\sqrt{b(2 a+b)} \leqslant \sqrt{a(2 a+b)} \\
+\sqrt{b(a+2 b)} \\
\Leftrightarrow a(a+2 b)+b(2 a+b) \leqslant a(2 a+b)+b(a+
\end{array}$$
$2 b)$
$$\Leftrightarrow 2 a b \leqslant a^{2}+b^{2} .$$
Next, p... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,342 |
Given $a>0, b>0$, prove
$$\sqrt{\frac{a}{3 a+b}}+\sqrt{\frac{b}{a+3 b}} \leqslant 1 \leqslant \sqrt{\frac{a}{a+3 b}}+\sqrt{\frac{b}{3 a+b}} .$$ | Proof: First, prove the left inequality, eliminate the denominator
$$\begin{array}{l}
\Leftrightarrow \sqrt{a(a+3 b)}+\sqrt{b(3 a+b)} \\
\leqslant \sqrt{(3 a+b)(a+3 b)} \\
\Leftrightarrow \sqrt{a b(a+3 b)(3 a+b)} \leqslant a^{2}+b^{2}+2 a b \\
\Leftrightarrow \sqrt{a b(a+3 b)(3 a+b)} \leqslant(a+b)^{2} . \\
\because \s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,343 |
Given $a>0, b>0, 0<\lambda \leqslant \frac{1}{2}$, prove
$$\left(a^{\lambda}+b^{\lambda}\right) \cdot\left[\frac{1}{(2 a+b)^{\lambda}}+\frac{1}{(a+2 b)^{\lambda}}\right] \leqslant \frac{4}{3^{\lambda}} \text {. }$$ | Proof: Given $a>0, b>0, 0<2 \lambda \leqslant 1$, using the power mean inequality, we get
$$\begin{array}{l}
\left(\frac{a}{2 a+b}\right)^{\lambda}+\left(\frac{b}{a+2 b}\right)^{\lambda} \\
=\left(\sqrt{\frac{a}{2 a+b}}\right)^{2 \lambda}+\left(\sqrt{\frac{b}{a+2 b}}\right)^{2 \lambda} \\
\leqslant 2^{1-2 \lambda}\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,344 |
Given $a>0, b>0$, prove that
$$\sqrt[3]{\frac{a}{2 a+b}}+\sqrt[3]{\frac{b}{a+2 b}} \leqslant \sqrt[3]{\frac{a}{a+2 b}}+\sqrt[3]{\frac{b}{2 a+b}}$$ | Proof: Let $\frac{b}{a}=x$, then the original inequality is equivalent to
$$\sqrt[3]{\frac{1}{2+x}}+\sqrt[3]{\frac{x}{1+2 x}} \leqslant \sqrt[3]{\frac{1}{1+2 x}}+\sqrt[3]{\frac{x}{2+x}}$$
Since
$$\begin{array}{l}
\left(\sqrt[3]{\frac{1}{2+x}}+\sqrt[3]{\frac{x}{1+2 x}}\right)-\left(\sqrt[3]{\frac{1}{1+2 x}}+\sqrt[3]{\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,345 |
Question 1 Given that $x, y, z$ are positive real numbers, prove:
$$\frac{x}{2 x+y+z}+\frac{y}{x+2 y+z}+\frac{z}{x+y+2 z} \leqslant$$
3
《In the process of transformation, the method used is substitution》 | Prove: For $a, b \in \mathbf{R}^{+}$, it is obvious that
$$\frac{1}{a+b} \leqslant \frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right).$$
Thus,
$$\begin{array}{l}
\frac{x}{2 x+y+z}+\frac{y}{x+2 y+z}+\frac{z}{x+y+2 z} \\
\leqslant \frac{x}{4}\left(\frac{1}{z+x}+\frac{1}{x+y}\right)+\frac{y}{4}\left(\frac{1}{x+y}+\frac{1}{y... | \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 736,347 |
Question 2 Given real numbers $a, b, c$ satisfying $a b c \neq 0$, prove:
$$\frac{a^{4}}{4 a^{4}+b^{4}+c^{4}}+\frac{b^{4}}{a^{4}+4 b^{4}+c^{4}}+\frac{c^{4}}{a^{4}+b^{4}+4 c^{4}} \leqslant \frac{1}{2}$$ | $$\begin{array}{l}
\frac{a^{4}}{4 a^{4}+b^{4}+c^{4}}+\frac{b^{4}}{a^{4}+4 b^{4}+c^{4}}+\frac{c^{4}}{a^{4}+b^{4}+4 c^{4}} \\
\leqslant \frac{a^{4}}{2 a^{4}+2 a^{2} b^{2}+2 c^{2} a^{2}}+\frac{b^{4}}{2 a^{2} b^{2}+2 b^{4}+2 b^{2} c^{2}} \\
\quad+\frac{c^{4}}{2 a^{2} c^{2}+2 b^{2} c^{2}+2 c^{4}} \\
=\frac{1}{2}\left(\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,348 |
Question 3 Given that $a, b, c \in \mathbf{R}^{+}$, prove:
$$\begin{array}{l}
\frac{a b}{a+b+2 c}+\frac{c b}{2 a+b+c}+\frac{c a}{a+2 b+c} \\
\leqslant \frac{1}{4}(a+b+c)
\end{array}$$ | $$\begin{array}{l}
\frac{a b}{a+b+2 c}=\frac{a b}{(c+a)+(b+c)} \\
\leqslant \frac{a b}{2 \sqrt{(c+a)(b+c)}} \leqslant \frac{a b}{4}\left(\frac{1}{c+a}+\frac{1}{b+c}\right), \\
\text { i.e., } \frac{a b}{a+b+2 c} \leqslant \frac{1}{4}\left(\frac{a b}{b+c}+\frac{a b}{c+a}\right) . \\
\text { Similarly, } \frac{b c}{2 a+b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,349 |
Question 7 Given $a, b, c \in \mathbf{R}^{+}$, prove:
$$\begin{array}{l}
\frac{a b}{a^{2}+b^{2}+2 c^{2}}+\frac{b c}{2 a^{2}+b^{2}+c^{2}}+\frac{c a}{a^{2}+2 b^{2}+c^{2}} \\
\leqslant \frac{3}{4}
\end{array}$$ | $$\begin{array}{l}
\frac{a b}{a^{2}+b^{2}+2 c^{2}}+\frac{b c}{2 a^{2}+b^{2}+c^{2}}+\frac{c a}{a^{2}+2 b^{2}+c^{2}} \\
=\frac{a b}{\left(b^{2}+c^{2}\right)+\left(c^{2}+a^{2}\right)}+\frac{b c}{\left(c^{2}+a^{2}\right)+\left(a^{2}+b^{2}\right)} \\
+\frac{c a}{\left(a^{2}+b^{2}\right)+\left(b^{2}+c^{2}\right)} \\
\leqslan... | \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 736,350 |
Deepen 1 Let $a, b, c$ be positive real numbers, $-2<\lambda<2$, prove:
$$\begin{aligned}
& \sqrt{a^{2}+\lambda a b+b^{2}}+\sqrt{b^{2}+\lambda b c+c^{2}}+\sqrt{c^{2}+\lambda c a+a^{2}} \\
\geqslant & \sqrt{2+\lambda}(a+b+c) .
\end{aligned}$$ | Prove that since $(a+b)^{2}=a^{2}+b^{2}+2 a b \leqslant a^{2}+b^{2}+a^{2}+b^{2}$,
so $a^{2}+b^{2} \geqslant \frac{1}{2}(a+b)^{2}$,
so $a^{2}+\lambda a b+b^{2}=\left(1-\frac{\lambda}{2}\right)\left(a^{2}+b^{2}\right)+$
$$\begin{array}{l}
\frac{\lambda}{2}(a+b)^{2} \\
\quad \geqslant\left(1-\frac{\lambda}{2}\right) \cdo... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,351 |
Deepen 2 Let $a, b, c$ be positive real numbers, $-2<\lambda<2$, prove:
$$\begin{aligned}
& \left(a^{2}+\lambda a b+b^{2}\right)\left(b^{2}+\lambda b c+c^{2}\right)\left(c^{2}+\lambda c a+a^{2}\right) \\
\geqslant & \left(\frac{2+\lambda}{3}\right)^{3}(a b+b c+c a)^{3} .
\end{aligned}$$ | Prove from the proof process of 1, we get
$$a^{2}+\lambda a b+b^{2} \geqslant \frac{2+\lambda}{4}(a+b)^{2}$$
Similarly,
$$\begin{array}{l}
b^{2}+\lambda b c+c^{2} \geqslant \frac{2+\lambda}{4}(b+c)^{2} \\
c^{2}+\lambda c a+a^{2} \geqslant \frac{2+\lambda}{4}(c+a)^{2}
\end{array}$$
Multiplying these three inequalities... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,352 |
Deepening 3 Let $a, b, c$ be positive real numbers, $-2<\lambda<2$, prove:
$$\begin{array}{l}
\sqrt{\left(a^{2}+\lambda a b+b^{2}\right)\left(b^{2}+\lambda b c+c^{2}\right)}+ \\
\sqrt{\left(b^{2}+\lambda b c+c^{2}\right)\left(c^{2}+\lambda c a+a^{2}\right)}+ \\
\sqrt{\left(c^{2}+\lambda c a+a^{2}\right)\left(a^{2}+\lam... | Prove that by implementing formula transformation, we get
$$a^{2}+\lambda a b+b^{2}=\frac{2-\lambda}{4}(a+b)^{2}+\frac{2+\lambda}{4}(a-b)^{2},$$
Thus, we can construct the complex numbers
$$\begin{array}{l}
z_{1}=\frac{\sqrt{2-\lambda}}{2}(a+b)+\frac{\sqrt{2+\lambda}}{2}(a-b) i, \\
z_{2}=\frac{\sqrt{2-\lambda}}{2}(b+c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,353 |
Problem 4-1 (2002 Vietnam Competition Question) Let real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=9$, prove that: $2(x+y+z)-x y z \leqslant 10$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Assume without loss of generality that $x^{2} \geqslant y^{2} \geqslant z^{2}$,
then we have $x^{2} \geqslant 3, 6 \geqslant y^{2}+z^{2} \geqslant 2 y z$.
By the Cauchy-Schwarz inequality, we get
$$\begin{array}{l}
{[2(x+y+z)-x y z]^{2}=[2(y+z)+x(2-y z)]^{2}} \\
\leqslant\left[(y+z)^{2}+x^{2}\right] \cdot\left[4+(2-y z... | null | Inequalities | proof | Yes | Yes | inequalities | false | 736,359 |
Question 4-2 Let $x, y, z$ be real numbers, and satisfy $x^{2}+y^{2}+z^{2}=r^{2}$ $(r>0)$, prove that: $x+y+z-\frac{2}{r^{2}} x y z \leqslant \sqrt{2} r$.
For the letter $r$ in question 4-2, taking special values, we get:
1. (1991 Polish Mathematical Competition) Let $x, y, z$ be real numbers, and satisfy $x^{2}+y^{2}... | Prove using the 2009-term AM-GM inequality, we get
$$\begin{array}{l}
x^{2009}+2008=x^{2009}+1^{2009}+1^{2009}+\cdots+1^{2009} \geqslant 2009 x, \\
\text { i.e., } x^{2009}-2008(x-1) \geqslant x .
\end{array}$$
Similarly, $y^{2009}-2008(y-1) \geqslant y$,
$$z^{2009}-2008(z-1) \geqslant z$$
Noting the common inequalit... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,360 |
Theorem 1 If $a, b, c$ are the lengths of the three sides of $\triangle A B C$, then
$$\begin{array}{l}
\sqrt{a b-b c+b^{2}}+\sqrt{c a-b c+c^{2}}>\sqrt{a b+c a+a^{2}} \\
\sqrt{b c-c a+c^{2}}+\sqrt{a b-c a+a^{2}}>\sqrt{a b+b c+b^{2}} \\
\sqrt{c a-a b+a^{2}}+\sqrt{b c-a b+b^{2}}>\sqrt{b c+c a+c^{2}}
\end{array}$$ | $$\begin{array}{l}
\text { Proof of Theorem } 1 \text { : If we denote the left side of the inequality to be proved as } P \text {, then we have } \\
P^{e}=a b+c a+(b-c)^{2}+2 \sqrt{a b-b c+b^{2}} \sqrt{c a-b c+c^{2}}= \\
a b+c a+(b-c)^{2}+2 \sqrt{\left[a^{2}-(b-c)^{2}\right] b c}
\end{array}$$
Thus, it suffices to pr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,361 |
Theorem 2 If $a, b, c \in R^{+}$, then
$$\frac{a b}{a^{2}+b^{2}+2 c^{2}}+\frac{b c}{2 a^{2}+b^{2}+c^{2}}+\frac{c a}{a^{2}+2 b^{2}+c^{2}} \leq \frac{3}{4} \text {. }$$ | Theorem 2 Proof: Applying the 2-variable mean inequality, we get
$$\begin{array}{l}
\frac{a b}{a^{2}+b^{2}+2 c^{2}}+\frac{b c}{2 a^{2}+b^{2}+c^{2}}+\frac{c a}{a^{2}+2 b^{2}+c^{2}}= \\
\frac{a b}{\left(b^{2}+c^{2}\right)+\left(c^{2}+a^{2}\right)}+\frac{b c}{\left(c^{2}+a^{2}\right)+\left(a^{2}+b^{2}\right)}+\frac{c a}{\... | \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 736,362 |
Theorem 3 Let the lengths of the two legs of a right triangle be $a$ and $b$, and the height on the hypotenuse $c$ be $h$, then $\frac{2 \sqrt{2}}{3}(c+h) \leq a+b<c+h$. | Theorem 3 Proof: Since $a=c \sin A, b=c \cos A, h=\frac{a b}{c}=c \sin A \cos A$, we have
$$\begin{array}{c}
F=\left(\frac{a+b}{c+h}\right)^{2}=\left(\frac{\sin A+\cos A}{1+\sin A \cos A}\right)^{2}= \\
\frac{1+\sin A \cos A}{1+2 \sin A \cos A+\sin ^{2} A \cos ^{2} A}= \\
1+\frac{\sin ^{2} A \cos ^{2} A}{1+2 \sin A \co... | \frac{2 \sqrt{2}}{3}(c+h) \leq a+b<c+h | Inequalities | proof | Yes | Yes | inequalities | false | 736,363 |
Theorem 4 Let $x, y, z \geq 0$ and $x^{2}+y^{2}+z^{2}=1$, then
$$\frac{x^{2}}{1-y z}+\frac{y^{2}}{1-z x}+\frac{z^{2}}{1-x y} \leq \frac{3}{2} \text { 。 }$$ | Theorem 4 Proof: Let $S=\frac{x^{2}}{1-y z}+\frac{y^{2}}{1-z x}+\frac{z^{2}}{1-x y}$. If $x=0$ (or $y=0$ or $z=0$), then $S=y^{2}+z^{2}=1<\frac{3}{2}$.
Next, we prove that $x y z \neq 0$, in which case $x, y, z \in(0,1)$. Notice that $\frac{x^{2}}{1-y z}=x^{2}+\frac{x^{2} y z}{1-y z}$, then we have
$$\begin{array}{l}
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,364 |
Theorem 6 If $x, y, z \in R^{+}, x+y+z=x y z$, then
$$x\left(1-y^{2}\right)\left(1-z^{2}\right)+y\left(1-z^{2}\right)\left(1-x^{2}\right)+z\left(1-x^{2}\right)\left(1-y^{2}\right) \geq 12 \sqrt{3}$$ | Theorem 6 Proof: Given the condition, $x, y, z \in R^{+}, x+y+z=x y z$, we have
$$x y z=x+y+z \geq 3 \sqrt[3]{x y z} \text {, i.e., } x y z \geq 3 \sqrt{3} \text { . }$$
From $x+y+z=x y z$, we can derive $x+y=z(x y-1)$, thus
$$\begin{array}{l}
x\left(1-y^{2}\right)\left(1-z^{2}\right)+y\left(1-z^{2}\right)\left(1-x^{2... | 12 \sqrt{3} | Inequalities | proof | Yes | Yes | inequalities | false | 736,366 |
Question 1 Let $x, y, z$ be positive real numbers, and satisfy $x+y+z=1$, prove that: $\frac{x}{x+y z}+\frac{y}{y+z x}+\frac{z}{z+x y} \leqslant \frac{9}{4}$. | Given the condition $x+y+z=1$, we know that $x+yz=x(x+y+z)+yz=(x+y)(z+x)$,
i.e., $x+yz=(x+y)(z+x)$.
Similarly, we get $y+zx=(y+z)(x+y)$,
$z+xy=(z+x)(y+z)$.
Thus, the inequality to be proven is equivalent to
$$\begin{array}{l}
\frac{x}{(x+y)(z+x)}+\frac{y}{(y+z)(x+y)}+ \\
\frac{z}{(z+x)(y+z)} \leqslant \frac{9}{4}, \\
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,367 |
Question 2 Let positive real numbers $x, y, z$ satisfy the relation $x+y+z=1$, prove that:
$$\sqrt{\frac{x}{x+y z}}+\sqrt{\frac{y}{y+z x}}+\sqrt{\frac{z}{z+x y}} \leqslant \frac{3 \sqrt{3}}{2} .$$ | $$\begin{array}{l}
\sqrt{\frac{x}{x+y z}}+\sqrt{\frac{y}{y+z x}}+\sqrt{\frac{z}{z+x y}} \leqslant \\
\sqrt{3\left(\frac{x}{x+y z}+\frac{y}{y+z x}+\frac{z}{z+x y}\right)}
\end{array}$$
Therefore, to prove $\sqrt{\frac{x}{x+y z}}+\sqrt{\frac{y}{y+z x}}+\sqrt{\frac{z}{z+x y}} \leqslant \frac{3 \sqrt{3}}{2}$, it suffices ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,368 |
Question 3 Let positive real numbers $x, y, z$ satisfy the relation $x+y+z=1$, prove:
$$\sqrt{\frac{y z}{x+y z}}+\sqrt{\frac{z x}{y+z x}}+\sqrt{\frac{x y}{z+x y}} \leqslant \frac{3}{2} .$$ | To prove that the inequality to be proved is equivalent to
$$\begin{array}{l}
\sqrt{\frac{y z}{(x+y)(z+x)}}+\sqrt{\frac{z x}{(y+z)(x+y)}}+ \\
\sqrt{\frac{x y}{(z+x)(y+z)}} \leqslant \frac{3}{2} .
\end{array}$$
By the two-variable mean inequality, we have
$$\begin{array}{l}
2 \sqrt{\frac{y z}{(x+y)(z+x)}}+2 \sqrt{\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,369 |
Question 4 Let positive real numbers $x, y, z$ satisfy the relation $x+y+z=1$, prove:
$$1<\frac{x-y z}{x+y z}+\frac{y-z x}{y+z x}+\frac{z-x y}{z+x y} \leqslant \frac{3}{2} .$$ | To prove the inequality is equivalent to
$$\frac{3}{4} \leqslant \frac{y z}{x+y z}+\frac{z x}{y+z x}+\frac{x y}{z+x y}<1$$
First, prove the right inequality:
$$\begin{array}{l}
\frac{y z}{x+y z}+\frac{z x}{y+z x}+\frac{x y}{z+x y} \\
=\frac{y z}{x(x+y+z)+y z}+\frac{z x}{y(x+y+z)+z x}+ \\
\frac{x y}{z(x+y+z)+x y}=\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,370 |
Question 5 Let positive real numbers $x, y, z$ satisfy the relation $x+y+z=1$, prove that: $\frac{1}{x+y z}+\frac{1}{y+z x}+\frac{1}{z+x y} \leqslant \frac{3}{4} \sqrt{\frac{3}{x y z}}$. | $$\begin{array}{l}
\text { Prove that the inequality to be proved is equivalent to } \\
\sum \frac{1}{x(x+y+z)+yz} \leqslant \frac{3}{4} \sqrt{\frac{3}{xyz(x+y+z)}} \\
\Leftrightarrow \sum \frac{1}{(x+y)(x+z)} \leqslant \frac{3}{4} \sqrt{\frac{3}{xyz(x+y+z)}} \\
\Leftrightarrow \sum(y+z) \leqslant \frac{3}{4} \sqrt{\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,371 |
Question 2 Given that $a, b, c$ are positive numbers, prove:
$$\frac{a^{2}+b c}{b+c}+\frac{b^{2}+c a}{c+a}+\frac{c^{2}+a b}{a+b} \geqslant a+b+c$$ | Prove that for positive numbers $x, y, z$,
$$\frac{z}{x y}(x-y)^{2}+\frac{x}{y z}(y-z)^{2}+\frac{y}{z x}(z-x)^{2} \geqslant 0,$$
By transformation, we can get $\frac{y z}{x}+\frac{z x}{y}+\frac{x y}{z} \geqslant x+y+z$
Let $x=b+c, y=c+a, z=a+b$, substitute into the inequality
(3), we get
$$\begin{array}{l}
\quad \frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,372 |
Given $a, b, c > 0$, and $abc = 1$, prove that $\frac{a}{a^{2}+2} + \frac{b}{b^{2}+2} + \frac{c}{c^{2}+2} \leqslant 1$. For this Baltic Olympiad problem, the author will provide the most elementary and simpler proof in this article, and ingeniously extend the problem-solving strategy to offer two excellent paired gener... | Proof: Let $A=(1+2a)(1+2b)+(1+2b)(1+2c)+(1+2c)(1+2a)=4(ab+bc+ca)+4(a+b+c)+3$
$B=(1+2a)(1+2b)(1+2c)=4(ab+bc+ca)+2(a+b+c)+9$
Thus, $A-B=2(a+b+c)-6$. Given $a, b, c > 0$ and $abc=1$, it follows that $a+b+c \geqslant 3\sqrt[3]{abc}=3$, so $A-B \geqslant 6-6=0$, which means $A \geqslant B > 0$. Also, $\frac{A}{B}=\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,374 |
Given $a_{i}>0, i=1,2, \cdots, n, n \in \mathrm{N}, n \geqslant 3$, and satisfying $a_{1} a_{2} \cdots a_{n}=1$. Then
$$\begin{array}{l}
\frac{1}{1+(n-1) a_{1}}+\frac{1}{1+(n-1) a_{2}}+\cdots+\frac{1}{1+(n-1) a_{n}} \\
\geqslant 1
\end{array}$$ | Prove by introducing parameter $\lambda$, construct the inequality
$$\frac{1}{1+(n-1) a_{1}} \geqslant \frac{a_{1}^{\lambda}}{a_{1}^{\lambda}+a_{2}^{\lambda}+\cdots+a_{n}^{\lambda}}$$
Below, we will study the feasibility of inequality (2).
Given $a_{1}>0, n \geqslant 3$, after rearrangement, inequality (2) is equivale... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,375 |
Problem 1 (1963 Moscow Mathematical Olympiad Question) Given that $a, b, c$ are positive numbers, prove: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}$. | Proof: Without loss of generality, assume $a \geqslant b \geqslant c$,
$$\begin{aligned}
\because & \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-\frac{3}{2} \\
= & \left(\frac{a}{b+c}-\frac{1}{2}\right)+\left(\frac{b}{c+a}-\frac{1}{2}\right)+\left(\frac{c}{a+b}-\frac{1}{2}\right) \\
= & \frac{2 a-b-c}{2(b+c)}+\frac{2 b-a-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,377 |
Proposition 1 Let $a, b, c \in(-2,+\infty)$, and $a^{2}+b^{2}+c^{2}+a b c=4$, then $a+b+c \leqslant 3$. (2) | Proof: By the Pigeonhole Principle, it is easy to see that among three real numbers $a, b, c$, at least two are not less than (or not greater than) 1. Without loss of generality, assume they are $b, c$, then $(b-1)(c-1) \geqslant 0$, i.e., $b+c \leqslant 1+bc$. (3)
From the condition, we get $4-a^{2}=b^{2}+c^{2}+abc \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,378 |
Proposition 2 Let real numbers $a, b, c$, satisfy $b+c>0, c+a>0, a+b>0$, and $a^{2}+b^{2}+c^{2}+a b c=4$, then $a+b+c>2$. (5) | Proof: (by contradiction) Assume $a+b+c \leqslant 2$,
then $00) \Rightarrow 2 a+b c \leqslant 0$. (6)
Similarly, $2 b+c a \leqslant 0$. (7)
(6) + (7), we get $2(a+b)+c(a+b) \leqslant 0$. (8)
Also, from the given conditions, it is easy to know that at least two of the numbers $a, b, c$ are positive. Without loss of ge... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,379 |
Proposition 3 Let $a, b, c \in \mathbf{R}^{+}$, and $a^{2}+b^{2}+c^{2}+a b c=4$, then $(b+c)^{2}+$
$$(c+a)^{2}+(a+b)^{2} \leqslant 12 \text {. (9) }$$ | Prove: Multiplying both sides of (3) by a positive number $a$, we get $a(b+c) \leqslant a + a b c$. (16)
Combining (11) and (4), we have $a b + c a + b c \leqslant 2 + a b c$, (11)
Multiplying (11) by 2, and adding $2\left(a^{2}+b^{2}+c^{2}\right)=2(4-a b c)$ (given), we immediately get $(b+c)^{2}+(c+a)^{2}+(a+b)^{2} \... | (b+c)^{2}+(c+a)^{2}+(a+b)^{2} \leqslant 12 | Inequalities | proof | Yes | Yes | inequalities | false | 736,380 |
(2011 Liaoning Province Liberal Arts College Entrance Examination Question 6) If the function $f(x)=\frac{x}{(2 x+1)(x-a)}$ is odd, then $a=(\quad)$.
(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) $\frac{3}{4}$
(D) 1 | Clever solution: Since the domain of an odd function is symmetric about the origin, and the domain of the function $f(x)=\frac{x}{(2 x+1)(x-a)}$ is $x \neq-\frac{1}{2}$ and $x \neq a$, it follows that $a=\frac{1}{2}$. Therefore, the correct choice is (A). | A | Algebra | MCQ | Yes | Yes | inequalities | false | 736,382 |
Given $a, b \in[1,3], a+b=4$,
prove: $\sqrt{10} \leqslant \sqrt{a+\frac{1}{a}}+\sqrt{b+\frac{1}{b}}<\frac{4 \sqrt{6}}{3}$. | By setting a variable, squaring, transforming into a univariate function, and using the monotonicity of the function to prove it.
Proof: Let \( y = \sqrt{a + \frac{1}{a}} + \sqrt{b + \frac{1}{b}} \), then
\[ y^2 = a + b + \frac{1}{a} + \frac{1}{b} + 2 \sqrt{\left(a + \frac{1}{a}\right)\left(b + \frac{1}{b}\right)} \]
... | \sqrt{10} \leq \sqrt{a + \frac{1}{a}} + \sqrt{b + \frac{1}{b}} < \frac{4 \sqrt{6}}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 736,383 |
Given: $a, b \in [1,3], a+b=4$, prove:
$$\left|\sqrt{a+\frac{1}{b}}-\sqrt{b+\frac{1}{a}}\right| \leqslant 2-\frac{2}{\sqrt{3}} .$$ | Prove: Let $y=\left|\sqrt{a+\frac{1}{b}}-\sqrt{b+\frac{1}{a}}\right|$, then
$y^{2}=a+b+\frac{1}{a}+\frac{1}{b}-2 \sqrt{\left(a+\frac{1}{b}\right)\left(b+\frac{1}{a}\right)}$
$=4+\frac{4}{a b}-2 \sqrt{a b+\frac{1}{a b}+2}$.
Notice the condition $a, b \in[1,3], a+b=4$, we get
$t=a b=a(4-a)=-a^{2}+4 a=-(a-2)^{2}+4 \in[3,4... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,384 |
Question 1 Given $x_{1}, x_{2} \in\left(0, \frac{\pi}{2}\right)$, prove: $\tan x_{1}+\tan x_{2} \geqslant 2 \tan \frac{x_{1}+x_{2}}{2}$. | Proof: Suppose $x_{1}, x_{2} \in \left(0, \frac{\pi}{2}\right)$, and $x_{1} \geqslant x_{2}$, then $\tan \frac{x_{1}+x_{2}}{2}>0, \tan \frac{x_{1}-x_{2}}{2} \geqslant 0$.
Thus, the original inequality is equivalent to
$\tan x_{1}-\tan \frac{x_{1}+x_{2}}{2} \geqslant \tan \frac{x_{1}+x_{2}}{2}-\tan x_{2}$,
which is $\t... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,385 |
Question 5 Let $a, b \in(0,1)$, prove:
$$\frac{a}{1-b^{2}}+\frac{b}{1-a^{2}} \geqslant \frac{a+b}{1-a b}$$ | Prove: Using the weighted sum of squares inequality
$$\frac{x^{2}}{p}+\frac{y^{2}}{q} \geqslant \frac{(x+y)^{2}}{p+q}, \quad p, q, x, y \in \mathbf{R}_{+} .$$
Thus, $\frac{a}{1-b^{2}}+\frac{b}{1-a^{2}}=\frac{a^{2}}{a-a b^{2}}+\frac{b^{2}}{b-a^{2} b}$
$$\begin{array}{l}
\geqslant \frac{(a+b)^{2}}{a+b-a b^{2}-a^{2} b} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,386 |
Question 6 Let $a, b, c \in(0,1)$, prove:
$$\frac{a}{1-a^{3}}+\frac{b}{1-b^{3}}+\frac{c}{1-c^{3}} \geqslant \frac{a+b+c}{1-a b c}$$ | Proof: First, we prove that $a^{4}+b^{4}+c^{4} \geqslant a b c(a+b+c)$. In fact, using the 4-term AM-GM inequality, we get
$$\begin{array}{l}
a b c(a+b+c) \\
=a \cdot a \cdot b \cdot c+a \cdot b \cdot b \cdot c+a \cdot b \cdot c \cdot c \\
\leqslant \frac{1}{4}\left(a^{4}+a^{4}+b^{4}+c^{4}\right)+\frac{1}{4}\left(a^{4}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,387 |
Question 7 Let $a, b, c \in(0,1)$, prove that:
$$\frac{a}{1-b^{2} c}+\frac{b}{1-c^{2} a}+\frac{c}{1-a^{2} b} \geqslant \frac{a+b+c}{1-a b c} .$$ | Proof: By adopting a similar method, we get
$$\begin{array}{l}
\frac{a}{1-b^{2} c}+\frac{b}{1-c^{2} a}+\frac{c}{1-a^{2} b} \\
=\frac{a^{2}}{a-a b^{2} c}+\frac{b^{2}}{b-a b c^{2}}+\frac{c^{2}}{c-a^{2} b c} \\
\geqslant \frac{(a+b+c)^{2}}{(a+b+c)-a b c(a+b+c)} \\
=\frac{a+b+c}{1-a b c}
\end{array}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,388 |
Question 8 Let $a, b, c, d \in(0,1)$, prove:
$$\frac{a}{1-a^{3} b}+\frac{b}{1-b^{3} c}+\frac{c}{1-c^{3} d}+\frac{d}{1-d^{3} a} \geqslant \frac{a+b+c+d}{1-a b c d}$$ | Prove: First, we prove that $a^{4} b+b^{4} c+c^{4} d+d^{4} a \geqslant a b c d(a+b+c+d)$.
Using the 51-element mean inequality, we get
$$\begin{array}{l}
\frac{23 a^{4} b+7 b^{4} c+11 c^{4} d+10 d^{4} a}{51} \geqslant \sqrt[51]{a^{102} b^{51} c^{51} d^{51}} \\
\text { i.e., } \frac{23 a^{4} b+7 b^{4} c+11 c^{4} d+10 d... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,389 |
Question 1: Given that $a, b, c$ are positive real numbers, prove:
$$\begin{array}{l}
\quad\left(a^{2}+b^{2}\right)^{2} \geqslant(a+b+c)(a+b-c)(b+c-a)(c \\
+a-b) .
\end{array}$$ | Proof: If among $a+b-c, b+c-a, c+a-b$, one of the values is zero, the inequality (1) obviously holds.
Below, we prove the case when $a+b-c, b+c-a, c+a-b$ are all non-zero.
(1) If among $a+b-c, b+c-a, c+a-b$, only one is positive, let's assume $a+b-c>0$. In this case, there should be $b+c-a<0$ and $c+a-b<0$. Adding the... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,393 |
Let $a, b$ be non-negative real numbers, prove that:
$$a^{3}+b^{3} \geqslant \sqrt{a b}\left(a^{2}+b^{2}\right) .$$ | Prove using the 3-variable mean inequality, we get
$a^{2} b+a b^{2}=a \cdot a \cdot b+a \cdot b \cdot b \leqslant \frac{a^{3}+a^{3}+b^{3}}{3}+\frac{a^{3}+b^{3}+b^{3}}{3}$,
which means $a^{2} b+a b^{2} \leqslant a^{3}+b^{3}$.
Using the 2-variable mean inequality, we get
$$\sqrt{a^{5} b}=\sqrt{a^{3} \cdot a^{2} b} \leqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,394 |
Question 2: Given $x, y, z \in \mathbf{R}, A, B, C$ are the three interior angles of a triangle, prove:
$$x^{2}+y^{2}+z^{2} \geqslant 2 y z \cos A+2 x z \cos B+$$
$2 x y \cos C$.
Equality holds if and only if $\frac{x}{\sin A}=\frac{y}{\sin B}=\frac{z}{\sin C}$. | Proof: Using the triangle angle sum theorem, and applying $2 p q \leqslant p^{2}+q^{2}$ twice, we get $2 y z \cos A+2 z x \cos B+2 x y \cos C=2 y z \cos [\pi-(B+C)]+2 x(z \cos B+$ $y \cos C) \leqslant-2 y z \cos (B+C)+x^{2}+(z \cos B+$ $y \cos C)^{2}=x^{2}+2 y z \sin B \sin C+z^{2} \cos ^{2} B+$ $y^{2} \cos ^{2} C \leq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,396 |
Question 3: Given that $a$, $b$, and $c$ are the lengths of the sides of an acute triangle, prove that: $b c \sqrt{b^{2}+c^{2}-a^{2}}+c a \sqrt{c^{2}+a^{2}-b^{2}}+$ $a b \sqrt{a^{2}+b^{2}-c^{2}}>2 \sqrt{2} a b c$. | $$\begin{array}{l}
\text{Prove: By transforming the inequality, we can obtain} \\
\sqrt{\frac{b^{2}+c^{2}-a^{2}}{2 a^{2}}}+\sqrt{\frac{c^{2}+a^{2}-b^{2}}{2 b^{2}}} \\
+\sqrt{\frac{a^{2}+b^{2}-c^{2}}{2 c^{2}}}>2 .
\end{array}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,397 |
Example 1 Given real numbers $x, y$ satisfy $x+y=1$, prove:
$$x y \leqslant \frac{1}{4}$$ | Proof: Since $x+y=1$, by the pigeonhole principle, it is known that $x, y$ are on opposite sides of $\frac{1}{2}$ (or at $\frac{1}{2}$),
i.e., $\left(x-\frac{1}{2}\right)\left(y-\frac{1}{2}\right) \leqslant 0$, thus $x y-\frac{1}{2}(x+$ $y)+\frac{1}{4} \leqslant 0 \Leftrightarrow x y \leqslant \frac{1}{2}(x+y)-\frac{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,398 |
Example 3 For any positive real numbers $a, b, c$, prove:
$$\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geqslant 9(a b+b c+c a)$$ | Prove: Let $\sum f(a, b, c)$ denote the cyclic sum over $a, b, c$.
By the principle of extraction, we know that $a^{2}, b^{2}, c^{2}$ have at least two on the same side of 1 (or at 1), without loss of generality, assume $a^{2}, b^{2}$ are on the same side of 1 (or at 1), then $\left(a^{2}-1\right)\left(b^{2}-1\right) ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,401 |
Example 4 For any positive real numbers $a, b, c$, prove:
$$\begin{array}{l}
\left(a^{5}-a^{2}+3\right)\left(b^{5}-b^{2}+3\right)\left(c^{5}-c^{2}+3\right) \geqslant(a+ \\
b+c)^{3}
\end{array}$$ | Prove that for any positive real number $x, x^{3}-1$ and $x^{2}-1$ have the same sign (or both are 0),
i.e., $0 \leqslant\left(x^{3}-1\right)\left(x^{2}-1\right)=x^{5}-x^{3}-x^{2}+1$ $=\left(x^{5}-x^{2}+3\right)-\left(x^{3}+2\right)$,
thus $\left(x^{5}-x^{2}+3\right) \geqslant\left(x^{3}+2\right)$.
Therefore, $\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,402 |
Proposition: For $\triangle A B C$ with sides $a, b, c$ and corresponding altitudes $h_{a}, h_{b}, h_{c}$, and circumradius $R$, then
$$\sum a^{2} \sum \frac{a^{2}}{h_{b}^{2}+h_{c}^{2}} \geqslant 18 R^{2}$$ | Proof: $\because h_{a}^{2}=\frac{4 \triangle^{2}}{a^{2}}, h_{b}^{2}=\frac{4 \triangle^{2}}{b^{2}}, h_{c}^{2}=$
$$\frac{4 \Delta^{2}}{c^{2}}, a^{2} b^{2} c^{2}=16 R^{2} \Delta^{2}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,403 |
Proposition: Let the internal angle bisectors of $\triangle A B C$ be $m_{a}$, $m_{b}$, and $m_{c}$. Then,
$$\frac{9}{4} \leqslant \frac{m_{a}^{2}}{b c}+\frac{m_{b}^{2}}{c a}+\frac{m_{c}^{2}}{a b} \leqslant \frac{R}{2 r}+\frac{r}{R}+\frac{3}{4} .$$ | Prove: From $m_{a}=\frac{\sqrt{2 b^{2}+2 c^{2}-a^{2}}}{2}$ we know
Leave the above text in its original format and line breaks, and output the translation directly. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,404 |
Example 1 Given $x+y+z=xyz$, prove:
$$\frac{x+y}{1-xy}+\frac{y+z}{1-yz}+\frac{z+x}{1-zx}=\frac{(x+y)(y+z)(z+x)}{(1-xy)(1-yz)(1-zx)}$$ | Staring at the fraction $\frac{x+y}{1-x y}$ in the proof target, from the given conditions, first transform this algebraic expression. In fact,
From $x+y+z=x y z$ we get $x+y=x y z-z$,
that is, $x+y=-z(1-x y)$,
so $\frac{x+y}{1-x y}=-z$.
Similarly, we have $\frac{y+z}{1-y z}=-x$,
$$\frac{z+x}{1-z x}=-y .$$
From the e... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,405 |
Example 2 Given $a(y+z)=x, b(z+x)=y$, $c(x+y)=z$, prove: $a b+b c+c a+2 a b c=1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Prove: Transform the given conditions into
$$a=\frac{x}{y+z}, b=\frac{y}{z+x}, c=\frac{z}{x+y},$$
Add 1 to both sides of the above three equations, then take the reciprocal, to get
$$\begin{array}{l}
\frac{1}{a+1}=\frac{y+z}{x+y+z}, \frac{1}{b+1}=\frac{z+x}{x+y+z}, \frac{1}{c+1} \\
=\frac{x+y}{x+y+z}
\end{array}$$
Ad... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,406 |
Example 3 Given $\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0$, prove:
$$\frac{a}{(b-c)^{2}}+\frac{b}{(c-a)^{2}}+\frac{c}{(a-b)^{2}}=0 .$$ | Prove: For the conditional equation $\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0$, multiply both sides by $\frac{1}{b-c}, \frac{1}{c-a}, \frac{1}{a-b}$ respectively, to get
$$\begin{array}{l}
\frac{a}{(b-c)^{2}}+\frac{b}{(b-c)(c-a)}+\frac{c}{(a-b)(b-c)}=0, \\
\frac{a}{(b-c)(c-a)}+\frac{b}{(c-a)^{2}}+\frac{c}{(c-a)(a-b)... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,407 |
Question 2 If $a, b$ are positive numbers, and $a+b=1$, then
$$\sqrt{a^{2}+1}+\sqrt{b^{2}+1}<1+\sqrt{2}$$ | Proof: $\because a b \leqslant\left(\frac{a+b}{2}\right)^{2}=\frac{1}{4} \Rightarrow a b \in\left(0, \frac{1}{4}\right)$, and $y=(x-1)^{2}+1$ is a decreasing function on $(-\infty, 1)$,
$$\begin{array}{l}
\therefore \sqrt{a^{2}+1}+\sqrt{b^{2}+1} \\
=\sqrt{\left(\sqrt{a^{2}+1}+\sqrt{\left.b^{2}+1\right)^{2}}\right.} \\
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,410 |
Question 3 Given $a, b \in \mathrm{R}^{+}, a+b=1$, prove
$$\sqrt{5} \leqslant \sqrt{1+a^{2}}+\sqrt{1+b^{2}}<1+\sqrt{2}$$ | Prove: In the right figure, $A B=B C=C D=1$, and $A B \perp B C, D C \perp B C$. Take any point $E$ on $B C$, let $B E=a$, then $C E=b, A E=\sqrt{1+a^{2}}, D E=\sqrt{1+b^{2}}$.
(1) In $\triangle A E D$, using the triangle inequality (the sum of any two sides is greater than the third side), we get
$$A E+E D \geqslant A... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,411 |
If $a, b \in \mathrm{R}^{+}$, then
$$\sqrt{a^{2}+1}+\sqrt{b^{2}+1}<1+\sqrt{(a+b)^{2}+1} \text {. }$$ | $$\begin{array}{l}
\Leftrightarrow a^{2}+b^{2}+2+2 \sqrt{\left(a^{2}+1\right)\left(b^{2}+1\right)}<2+a^{2}+b^{2} \\
+2 a b+2 \sqrt{(a+b)^{2}+1} \\
\quad \Leftrightarrow \sqrt{\left(a^{2}+1\right)\left(b^{2}+1\right)}<a b+\sqrt{(a+b)^{2}+1} \\
\quad \Leftrightarrow a^{2} b^{2}+a^{2}+b^{2}+1<a^{2} b^{2}+a^{2}+b^{2}+2 a b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,412 |
Question 7 If $a_{1}, a_{2}, \cdots, a_{n} \in \mathrm{R}^{+}, n>1$, then
$$\sum_{i=1}^{n} \sqrt{a_{i}^{2}+1}<n-1+\sqrt{\left(\sum_{i=1}^{n} a_{i}\right)^{2}+1}$$ | Proof: (1) When $n=2$, it has been proven in the previous text;
(2) Assume that when $n=k(2 \leqslant k \in \mathrm{N})$, the inequality holds:
$$\sum_{i=1}^{k} \sqrt{a_{i}^{2}+1}<k-1+\sqrt{\left(\sum_{i=1}^{k} a_{i}\right)^{2}+1} \text {, }$$
Then, when $n=k+1$,
$$\begin{array}{l}
\sum_{i=1}^{k+1} \sqrt{a_{i}^{2}+1}=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,413 |
Question 8 If $a_{1}, a_{2}, \cdots, a_{n} \in \mathrm{R}^{+}$, and $\sum_{i=1}^{N} a_{i}=1$, then $\sum_{i=1}^{n} \sqrt{a_{i}+1}>n-1+\sqrt{2}$. | Proof: From the given, we have
$$
0\frac{a_{1}}{\sqrt{1+1}+1} \\
=(\sqrt{2}-1) a_{1}
\end{array}$$
Thus, $\sqrt{a_{t}+1}>1+(\sqrt{2}-1) a_{1}$. Therefore, $\sum_{t=1}^{n} \sqrt{a_{t}+1}$
$$>\sum_{i=1}^{n}\left[1+(\sqrt{2}-1) a_{i}\right]=n+(\sqrt{2}-1) \sum_{i=1}^{n} a_{i}=n-1+\sqrt{2}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,414 |
Question 3 If $n$ is an integer greater than 2, then for any right triangle, with the lengths of the legs being $a$ and $b$, and the length of the hypotenuse being $c$, we have $a^{n}+b^{n}<c^{n}$. | Prove: Since $a < c$ and $b < c$, it follows that $a^{n-2} < c^{n-2}$, $b^{n-2} < c^{n-2}$. And note that $a^{2} + b^{2} = c^{2}$, thus we have
$$\begin{aligned}
a^{n} & +b^{n}=a^{2} \cdot a^{n-2}+b^{2} \cdot b^{n-2} \\
& <a^{2} \cdot c^{n-2}+b^{2} \cdot c^{n-2} \\
= & \left(a^{2}+b^{2}\right) c^{n-2}=c^{2} \cdot c^{n-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,416 |
Question 4 In $\triangle A B C$, the sides opposite to angles $A, B, C$ are $a, b, c$ respectively, and $a^{n}+b^{n}=c^{n}(n \geqslant 3)$. Then the shape of $\triangle A B C$ is $(\quad)$.
A. Right triangle
B. Acute triangle
C. Obtuse triangle
D. Right or obtuse triangle | Solution: Clearly, $c$ is the largest side. Since $\left(\frac{a}{c}\right)^{n}+\left(\frac{b}{c}\right)^{n}=1$ ($n \geqslant 3$), we have
$$\left(\frac{a}{c}\right)^{2}+\left(\frac{b}{c}\right)^{2}>\left(\frac{a}{c}\right)^{n}+\left(\frac{b}{c}\right)^{n}=1$$
Thus, $a^{2}+b^{2}>c^{2}$. By the cosine rule, the largest... | B | Geometry | MCQ | Yes | Yes | inequalities | false | 736,417 |
Question 1 As shown in Figure 1, find the maximum value of the area of the rectangle inscribed in the semicircle $O$ (the radius of the circle is 1).
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Parse: Connect $O A$.
Figure 1
Consider from the perspective of trigonometric functions. Let $\angle A O B=\theta$ $\left(0<\theta<\frac{\pi}{2}\right)$, then
$$|A B|=\sin \theta,|B C|=2|O B|=2 \cos \theta,$$
Thus, the area of rectangle $A B C D$ is
$$S=|A B| \cdot|B C|=2 \sin \theta \cos \theta=\sin 2 \theta$$
When ... | 1 | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,418 |
Question 2 As shown in Figure 2, find the maximum value of the area of an inscribed isosceles trapezoid in a semicircle (the radius of the circle is 1).
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Analysis: From an algebraic perspective. Let
Figure 2 $O E=x(0<x<1)$, draw $C E \perp A B$, with the foot of the perpendicular being $E$, then the area of the isosceles trapezoid $A B C D$ is
$$S=\frac{1}{2}(|A B|+|C D|) \cdot|C E|,$$
which is $S=(1+x) \sqrt{1-x^{2}}$.
The extremum of this function can be found using ... | \frac{3 \sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,419 |
Question 3 As shown in Figure 3, given the diameter $A B=8 \mathrm{~cm}$ of circle $O$, chord $A D=C D$ $=2 \mathrm{~cm}$, find the length of $B C$. | Connect $AC, BD$. In the right-angled triangle, by the Pythagorean theorem, we get
$$BD=\sqrt{AB^{2}-AD^{2}}=\sqrt{8^{2}-2^{2}}=2\sqrt{15}$$
By the acute angle trigonometric functions, we get
$$\sin \angle DAB=\frac{BD}{AB}=\frac{2\sqrt{15}}{8}=\frac{\sqrt{15}}{4}$$
In the right-angled $\triangle ABC$, by the acute a... | 7 | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,420 |
Question 5 As shown in Figure 6, find the maximum area of an arbitrary quadrilateral $ABCD$ inscribed in a semicircle $O$ with radius 1. | Parse: Connect $O C, O D$, let
$$\angle B O C=\alpha, \angle C O D=\beta, \angle D O A=\gamma$$
Obviously $\alpha+\beta+\gamma=\pi$, then the area of quadrilateral $A B C D$
$$S=\frac{1}{2}(\sin \alpha+\sin \beta+\sin \gamma)$$
By the common inequality $\sin \alpha+\sin \beta+\sin \gamma \leqslant \frac{3 \sqrt{3}}{2... | \frac{3 \sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,422 |
Question 6 As shown in Figure 7, there is a semi-elliptical steel plate, with the length of the major semi-axis being $2 r$ and the length of the minor semi-axis being $r$. It is planned to cut this steel plate into the shape of an isosceles trapezoid, with the lower base $A B$ being the minor axis of the semi-ellipse,... | (1) According to the problem, establish a rectangular coordinate system $xOy$ with the midpoint $O$ of $AB$ as the origin (Figure 8). Clearly, point $C(x, y)$ satisfies the equation $\frac{x^{2}}{r^{2}} + \frac{y^{2}}{4 r^{2}} = 1 (y \geqslant 0)$,
Solving for $y$ gives $y = 2 \sqrt{r^{2} - x^{2}} (0 < x < r)$.
Let $... | \frac{3 \sqrt{3}}{2} r^{2} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,423 |
Question 7 As shown in Figure 9, the three sides $AB$, $BC$, and $CD$ of the isosceles trapezoid $ABCD$ are tangent to the graph of the function $y=-\frac{1}{2} x^{2}+2, x \in[-2,2]$ at points $P$, $Q$, and $R$. Find the minimum value of the area of trapezoid $ABCD$. | From the appearance of the graph, question 6 and question 7 have very similar aspects. Here is the solution: $\square$
Let the area of the isosceles trapezoid $A B C D$ be $S$, and the coordinates of point $P$ be $\left(t, -\frac{1}{2} t^{2} + 2\right) (0 < t \leqslant 2)$. According to the problem, the coordinates of... | 4 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,424 |
Question 1 In the "Mathematical Problems and Solutions" column of Issue 2, 2003 of "Mathematics Teaching", Problem 580 is as follows:
Let $a, b, c$ be the sides of $\triangle ABC$, prove that:
$$\frac{a}{2a+b-c}+\frac{b}{2b+c-a}+\frac{c}{2c+a-b} \geqslant \frac{3}{2}.$$
The author attempted to explore the upper bound... | Proof: First, prove the right inequality,
$$\begin{array}{l}
\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3} \leqslant \frac{3}{4} \\
\Leftrightarrow 4(b+3)(c+3)+4(c+3)(a+3)+ \\
4(a+3)(b+3) \leqslant 3(a+3)(b+3)(c+3) \\
\Leftrightarrow 5(a b+b c+c a)+3(a+b+c) \geqslant 24 .
\end{array}$$
Noting that $a b c=1$, we apply the... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,426 |
Question 2.2 Given $x, y, z \in \mathrm{R}_{+}, k>1$, prove:
$$\frac{x}{x+k y+z}+\frac{y}{x+y+k z}+\frac{z}{k x+y+z} \geqslant \frac{3}{k+2} \text {. }$$ | Prove: Introduce the transformation
$$\left\{\begin{array}{l}
k x+y+z=a, \\
x+k y+z=b, \\
x+y+k z=c .
\end{array}\right.$$
Solve for
$$\left\{\begin{array}{l}
x=\frac{1}{(k-1)(k+2)}[(k+1) a-b-c], \\
y=\frac{1}{(k-1)(k+2)}[(k+1) b-c-a], \\
z=\frac{1}{(k-1)(k+2)}[(k+1) c-a-b] .
\end{array}\right.$$
Thus, $\frac{x}{x+k ... | \frac{3}{k+2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,427 |
Question 2.1 Given $x, y, z \in \mathrm{R}_{+}, k>1$, prove:
$$\frac{x}{k x+y+z}+\frac{y}{x+k y+z}+\frac{z}{x+y+k z} \leqslant \frac{3}{k+2} .$$ | Prove: Introducing transformation is too troublesome, equivalent conversion to $\geq$
$$\left\{\begin{array}{l}
k x+y+z=a, \text { then use Cauchy-Schwarz and it's done } \\
x+k y+z=b, \\
x+y+k z=c .
\end{array}\right.$$
Solve for
$$\left\{\begin{array}{l}
x=\frac{1}{(k-1)(k+2)}[(k+1) a-b-c], \\
y=\frac{1}{(k-1)(k+2)}... | \frac{3}{k+2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,428 |
Question 2 Let $a, b>0$, and $a+b=1$, prove:
$$\frac{a}{1+a}+\frac{b}{1+b}<\frac{1}{1+a b}$$ | Prove the inequality (2) $\Leftrightarrow \frac{a+b+2 a b}{1+a+b+a b}<\frac{1}{1+a b}$
$$\begin{array}{l}
\Leftrightarrow \frac{1+2 a b}{2+a b}<\frac{1}{1+a b} \\
\Leftrightarrow(1+2 a b)(1+a b)<2+a b \\
\Leftrightarrow 1+3 a b+2 a^{2} b^{2}<2+a b \\
\Leftrightarrow 2 a b(1+a b)<1
\end{array}$$
Notice that $a b \leqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,429 |
Question 5 Let $a, b>0$, and $a+b=1$, prove:
$$\frac{a}{1+a}+\frac{b}{1+b} \leqslant \frac{5}{6(1+ab)}$$ | Prove using the AM-GM inequality, on one hand, we have
$$\begin{aligned}
& \frac{a}{1+a}+\frac{b}{1+b}=1-\frac{1}{1+a}+1-\frac{1}{1+b} \\
= & 2-\left(\frac{1}{1+a}+\frac{1}{1+b}\right) \\
= & 2-\frac{1}{3}[(1+a)+(1+b)] \cdot\left(\frac{1}{1+a}+\frac{1}{1+b}\right) \\
\leqslant & 2-\frac{1}{3} \cdot 2 \sqrt{(1+a)(1+b)} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,431 |
Question 7 Let $x_{i}>0, i=1,2, \cdots, n(n \geqslant 2)$, and $x_{1}+x_{2}+\cdots+x_{n}=1$, prove:
$$\begin{array}{l}
\frac{x_{1}}{1+x_{1}}+\frac{x_{2}}{1+x_{2}}+\cdots+\frac{x_{n}}{1+x_{n}} \\
\leqslant \frac{n^{n}+1}{n^{n-1}(n+1)\left(1+x_{1} x_{2} \cdots x_{n}\right)}
\end{array}$$ | Prove that on the one hand, applying the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& \frac{x_{1}}{1+x_{1}}+\frac{x_{2}}{1+x_{2}}+\cdots+\frac{x_{n}}{1+x_{n}} \\
= & n-\left(\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{n}}\right) \\
= & n-\frac{1}{n+1} \cdot\left[\left(1+x_{1}\right)+\left(1+x_{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,432 |
Question 8 Let $x_{i}>0, i=1,2, \cdots, n(n \geqslant 2)$, and $x_{1}+x_{2}+\cdots+x_{n}=1$, prove:
$$\frac{x_{1}}{1+x_{1}}+\frac{x_{2}}{1+x_{2}}+\cdots+\frac{x_{n}}{1+x_{n}} \leqslant \frac{1}{1+\sqrt{x_{1} x_{2} \cdots x_{n}}}$$ | Prove that the function $f(x)=\frac{x}{1+x}$ is a convex function on $(0,+\infty)$.
On one hand, applying Jensen's inequality, we have:
$$\begin{aligned}
& \frac{x_{1}}{1+x_{1}}+\frac{x_{2}}{1+x_{2}}+\cdots+\frac{x_{n}}{1+x_{n}} \\
\leqslant & n \cdot \frac{\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}}{1+\frac{x_{1}+x_{2}+\cdot... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,433 |
Example 1 Given $x, y$ satisfy $x^{2}+y^{2}-4 x-6 y+9=0$, prove: $19 \leqslant x^{2}+y^{2}+12 x+6 y \leqslant 99$. | Analyze and prove the conditional equation
$$x^{2}+y^{2}-4 x-6 y+9=0$$
This is the general form of the equation of a circle, which can be transformed into the standard form
$$(x-2)^{2}+(y-3)^{2}=2^{2} \text{ (completing the square) }$$
To prove \(19 \leqslant x^{2}+y^{2}+12 x+6 y \leqslant 99\), from (1) we get \(x^{... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,434 |
Example 2 Given $x, y, z \in \mathbf{R}_{+}$, and $x+y+z=1$, prove: $\frac{1}{x}+\frac{4}{y}+\frac{9}{z} \geqslant 36$. | Analyzing and proving the inequality $\frac{1}{x}+\frac{4}{y}+\frac{9}{z} \geqslant 36$, we observe that the denominators are linear in the variables, while the numerators are constants, including the factor 1. This can be handled by using the reverse of the given condition $1=x+y+z$, i.e., substituting 1 with $x+y+z$.... | 36 | Inequalities | proof | Yes | Yes | inequalities | false | 736,435 |
Example 3 Given that $a, b, c$ are positive real numbers, and $a b c + a + c = b$, try to prove:
$$\frac{2}{a^{2}+1}-\frac{2}{b^{2}+1}+\frac{3}{c^{2}+1} \leqslant \frac{10}{3}$$ | The inequality to be proved is a conditional inequality with three variables, which can be simplified by the conditional equation $abc + a + c = b$, solving for $c = \frac{b-a}{1+ab}$, and substituting it into the inequality to be proved, thus reducing it to an inequality with two variables, which serves to eliminate v... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,436 |
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