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Example 4 Given that $a, b, c$ are positive real numbers, and $ab + bc + ca = 1$, try to prove: $\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}+\frac{1}{c^{2}+1} \leqslant \frac{9}{4}$. | It is quite interesting that, by replacing the 1 in the denominator with $a b + b c + c a$, we can factorize and transform, getting $a^{2} + 1 = a^{2} + a b + b c + c a = (a + b)(a + c)$ (factorization),
i.e., $a^{2} + 1 = (c + a)(a + b)$.
Similarly, we get $b^{2} + 1 = (a + b)(b + c)$ (symmetric thinking),
$$c^{2} + ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,437 |
Example 5 Given that $a, b, c$ are positive real numbers, and satisfy
$$\frac{a^{2}}{a^{2}+1}+\frac{b^{2}}{b^{2}+1}+\frac{c^{2}}{c^{2}+1}=1$$
Prove: $a b c \leqslant \frac{\sqrt{2}}{4}$. | Analysis and Proof If we regard each fraction in the conditional equation as a letter and perform local substitution, we can let
$$x=\frac{a^{2}}{a^{2}+1}, y=\frac{b^{2}}{b^{2}+1}, z=\frac{c^{2}}{c^{2}+1},$$
then $x, y, z \in(0,1), x+y+z=1$, and
$$a=\sqrt{\frac{x}{1-x}}, b=\sqrt{\frac{y}{1-y}}, c=\sqrt{\frac{z}{1-z}} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,438 |
Example 6 is the same as Example 5.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Analyze and Prove: Carefully read the trigonometric substitution method in the original publication. Its core involves rearranging equations and using the AM-GM inequality for two variables, similarly deriving two inequalities, and finally multiplying the three inequalities of the same direction to obtain the proof. In... | a b c \leqslant \frac{\sqrt{2}}{4} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,439 |
$$\begin{aligned}
& \text { Lemma If } f(x)=\frac{1}{1+x^{k}}, x \in(0,1), n \in N^{*}, n \geq k+1, k \geq 2 \text {, then } \\
f(x) \leq & \frac{k n^{k+1}}{\left(1+n^{k}\right)^{2}} x+\frac{n^{k}\left(n^{k}+k+1\right)}{\left(1+n^{k}\right)^{2}}
\end{aligned}$$
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 $k n^{k+1} x^{k+1}-n^{k}\left(n^{k}+k+1\right) x^{k}+k n^{k+1} x-(k-1) n^{k}+1 \leq 0$.
Let $g(x)=k n^{k+1} x^{k+1}-n^{k}\left(n^{k}+k+1\right) x^{k}+k n^{k+1} x-(k-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,440 |
Example 1 Let the three sides of $\triangle A B C$ be $a, b, c$, and the area be $\Delta$, then $\min \left\{\sqrt{a^{4}+b^{4}}, \sqrt{b^{4}+c^{4}}, \sqrt{c^{4}+a^{4}}\right\} \geqslant 2 \sqrt{2} \Delta$. | Proof: Without loss of generality, let $a \geqslant b \geqslant c$. Then, by the two-variable mean inequality and the boundedness of the sine function, we get
$$\begin{array}{l}
\min \left\{\sqrt{a^{4}+b^{4}}, \sqrt{b^{4}+c^{4}}, \sqrt{c^{4}+a^{4}}\right\} \geqslant \sqrt{b^{4}+c^{4}} \\
\geqslant \sqrt{2 b^{2} c^{2}} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,441 |
Example 2 Given $x, y, z \in \mathbf{R}_{+}, x^{2}+y^{2}+z^{2} \leqslant \sqrt{3} x y z$, prove: $x+y+z \leqslant x y z$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $$\begin{aligned}
& \text { Prove: from } x^{2}+y^{2} \geqslant 2 x y, \\
& 2\left(a^{2}+b^{2}+c^{2}\right)=\left(a^{2}+b^{2}\right)+\left(b^{2}+c^{2}\right)+\left(c^{2}+a^{2}\right) \\
\geqslant & 2 a b+2 b c+2 c a,
\end{aligned}$$
Therefore, $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$.
Thus, we have $(a+b+c)^{2}=a^{2}... | null | Inequalities | proof | Yes | Yes | inequalities | false | 736,442 |
Example 11 If $a, b \in \mathbf{R}_{+}, a+b=1$, prove:
(1) $\frac{a}{a^{2}+b}+\frac{b}{a+b^{2}} \leqslant \frac{4}{3}$;
(2) $\frac{a}{a+b^{3}}+\frac{b}{a^{3}+b} \leqslant \frac{8}{5}$. | Proof: Since $a b \leqslant\left(\frac{a+b}{2}\right)^{2}=\frac{1}{4}$, let $t=a b$, then we have $t \in\left(0, \frac{1}{4}\right]$.
(1) From $a+b=1$, we get
$a^{2}+b^{2}=1-2 a b, a^{3}+b^{3}=1-3 a b$. Therefore,
$$\begin{array}{l}
\frac{a}{a^{2}+b}+\frac{b}{a+b^{2}} \leqslant \frac{4}{3} \\
\Leftrightarrow 3 a\left(a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,443 |
Example 12 If $a, b, c$ are positive numbers, $a+b+c=3$, prove:
$$\left(\frac{3}{a}-2\right)\left(\frac{3}{b}-2\right)\left(\frac{3}{c}-2\right) \leqslant 1$$ | Proof: Obviously, the inequality to be proved is equivalent to
$(3-2 a)(3-2 b)(3-2 c) \leqslant a b c$.
From $a+b+c=3$, we get $3-a=b+c, 3-b=c+a, 3-c=a+b$. Therefore, the inequality to be proved is equivalent to
$(b+c-a)(c+a-b)(a+b-c) \leqslant a b c . \quad(* *)$
If any of $b+c-a, c+a-b, a+b-c$ is zero, the inequality... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,444 |
Example 13 Given that $x, y, z$ are positive real numbers, prove:
$$\frac{x}{2 x+y+z}+\frac{y}{x+2 y+z}+\frac{z}{x+y+2 z} \leqslant \frac{3}{4} .$$ | $$\begin{array}{l}
\frac{x}{2 x+y+z}+\frac{y}{x+2 y+z}+\frac{z}{x+y+2 z} \\
=\frac{x}{(z+x)+(x+y)}+\frac{y}{(x+y)+(y+z)} \\
+\frac{z}{(y+z)+(z+x)} \\
\leqslant \frac{x}{4}\left(\frac{1}{z+x}+\frac{1}{x+y}\right)+\frac{y}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right) \\
+\frac{z}{4}\left(\frac{1}{y+z}+\frac{1}{z+x}\right) ... | \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 736,445 |
Example 14 Given that if $a, b, c \in \mathbf{R}_{+}$, prove:
$$\frac{a b}{a+b+2 c}+\frac{b c}{2 a+b+c}+\frac{c a}{a+2 b+c} \leqslant \frac{1}{4}(a+b+c) .$$ | Prove: Applying the binary mean inequality, we get
$$\frac{a b}{a+b+2 c}=\frac{a b}{(c+a)+(b+c)} \leqslant \frac{a b}{4}\left(\frac{1}{c+a}+\frac{1}{b+c}\right),$$
i.e., $\frac{a b}{a+b+2 c} \leqslant \frac{1}{4}\left(\frac{a b}{b+c}+\frac{a b}{c+a}\right)$.
Similarly, $\frac{b c}{2 a+b+c} \leqslant \frac{1}{4}\left(\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,446 |
Example 15 Let $x, y, z \in \mathbf{R}_{+}$, prove:
$$\sqrt{x^{2}+y^{2}}+\sqrt{y^{2}+z^{2}}+\sqrt{z^{2}+x^{2}} \geqslant \sqrt{2}(x+y+z) .$$ | Prove: From $\frac{x^{2}+y^{2}}{2} \geqslant\left(\frac{x+y}{2}\right)^{2}$, taking the square root, and rearranging, we get
$$\sqrt{x^{2}+y^{2}} \geqslant \frac{\sqrt{2}}{2}(x+y)$$
Similarly, $\sqrt{y^{2}+z^{2}} \geqslant \frac{\sqrt{2}}{2}(y+z)$,
$$\sqrt{z^{2}+x^{2}} \geqslant \frac{\sqrt{2}}{2}(z+x)$$
Adding the t... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,447 |
Example 16 Let $a, b, c \in \mathbf{R}_{+}, abc=1$, prove:
$$\begin{array}{l}
\frac{1}{(a+1)^{2}+\sqrt{2\left(b^{4}+1\right)}}+\frac{1}{(b+1)^{2}+\sqrt{2\left(c^{4}+1\right)}} \\
+\frac{1}{(c+1)^{2}+\sqrt{2\left(a^{4}+1\right)}} \leqslant \frac{1}{2}
\end{array}$$ | Proof: From the simple inequality $\frac{x^{4}+y^{4}}{2} \geqslant\left(\frac{x^{2}+y^{2}}{2}\right)^{2}$, we get $\sqrt{2\left(x^{4}+y^{4}\right)} \geqslant x^{2}+y^{2}$, thus
$(a+1)^{2}+\sqrt{2\left(b^{4}+1\right)} \geqslant(a+1)^{2}+b^{2}+1=a^{2}$ $+b^{2}+2 a+2 \geqslant 2 a b+2 a+2$,
which means $(a+1)^{2}+\sqrt{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,448 |
Example 1 (2007 College Entrance Examination Mathematics, Shaanxi Volume, Science Question 19) As shown in Figure 1, in the quadrilateral pyramid $P-AB-$ $CD$, $AD \parallel BC, \angle ABC=90^{\circ}$, $PA \perp$ plane $ABCD, PA=4, AD$ $=2, AB=2 \sqrt{3}, BC=6$. (I) Prove: $BD \perp$ plane $PAC$;
(II) Find the size of ... | Solution: (I) Omitted.
(II) Let $B D \cap A C$ at $E$. From (I), we know $D E \perp$ plane $P A C$. Draw $E F \perp P C$ from $E$, with the foot of the perpendicular being $F$. Connect $D F$. By the theorem of three perpendiculars, we know $P C \perp D F$, thus $\angle E F D$ is the plane angle of the dihedral angle $A... | \arctan \frac{2 \sqrt{3}}{9} | Geometry | proof | Yes | Yes | inequalities | false | 736,449 |
Example 3 Let $x, y, z \in \mathbf{R}_{+}$, prove that:
$$\begin{array}{l}
\quad \sqrt{x^{2}+x y+y^{2}}+\sqrt{y^{2}+y z+z^{2}}+\sqrt{z^{2}+z x+x^{2}} \\
\geqslant \sqrt{3}(x+y+z) .
\end{array}$$ | Proof: From $a^{2}+b^{2} \geqslant 2 a b$, we get
$$\begin{array}{l}
2 \sqrt{x^{2}+x y+y^{2}}=\sqrt{3\left(x^{2}+y^{2}\right)+\left(x^{2}+y^{2}\right)+4 x y} \\
\geqslant \sqrt{3\left(x^{2}+y^{2}\right)+2 x y+4 x y}=\sqrt{3}(x+y), \\
\text { i.e., } \sqrt{x^{2}+x y+y^{2}} \geqslant \frac{\sqrt{3}}{2}(x+y) .
\end{array}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,450 |
Example 4 Given $x, y, z \in \mathbf{R}, A, B, C$ are the three interior angles of a triangle, prove:
$$x^{2}+y^{2}+z^{2} \geqslant 2 y z \cos A+2 z x \cos B+2 x y \cos C .$$
Equality holds if and only if $\frac{x}{\sin A}=\frac{y}{\sin B}=\frac{z}{\sin C}$. | Proof: Using the sum of the interior angles of a triangle and applying $2 p q \leqslant p^{2}+q^{2}$ twice, we get
$$\begin{array}{l}
2 y z \cos A+2 z x \cos B+2 x y \cos C \\
=2 y z \cos [\pi-(B+C)]+2 x(z \cos B+2 y \cos C) \\
\leqslant-2 y z \cos (B+C)+x^{2}+(z \cos B+2 y \cos C)^{2} \\
=x^{2}+2 y z \sin B \sin C+z^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,451 |
Example 5 Given $x, y \in \mathbf{R}$, and $x \neq y$, prove: $\left|\frac{1}{1+x^{2}}-\frac{1}{1+y^{2}}\right|<|x-y|$. | Prove: By applying the technique of finding a common denominator to the algebraic expression inside the absolute value on the left side of the inequality, the result on the right side of the inequality can be decomposed and isolated.
$$\begin{array}{l}
\left|\frac{1}{1+x^{2}}-\frac{1}{1+y^{2}}\right| \\
=|x-y| \cdot\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,452 |
Example 6 Let positive real numbers $x, y$ satisfy $x^{3}+y^{3}=x-y$, prove that $x^{2}+4 y^{2}<1$ | Proof: By the binary mean inequality, we get $5 y^{3}+x^{2} y \geqslant 2 \sqrt{5 x^{2} y^{4}}>4 x y^{2}$, which means $x^{3}-4 y^{3}-x^{2} y+4 x y^{2}<x^{3}+y^{3}$, $\left(x^{2}+4 y^{2}\right)(x-y)<x^{3}+y^{3}$, thus $x^{2}+4 y^{2}<\frac{x^{3}+y^{3}}{x-y}=1$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,453 |
Example 7 Let $a, b, c$ be positive numbers, and $a+b+c \geqslant a b c$, prove: $a^{2}+b^{2}+c^{2} \geqslant \sqrt{3} a b c$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Proof: The condition $a+b+c \geqslant a b c$ can be transformed into
$$\frac{1}{b c}+\frac{1}{c a}+\frac{1}{a b} \geqslant 1$$
The conclusion $a^{2}+b^{2}+c^{2} \geqslant \sqrt{3} a b c$ can be transformed into
$$\frac{a}{b c}+\frac{b}{c a}+\frac{c}{a b} \geqslant \sqrt{3}$$
In fact, since $\frac{a}{b c}+\frac{b}{c a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,454 |
Example 8 Given $x, y, z \in \mathbf{R}_{+}$, prove: $\sqrt{\frac{x}{y+z}}+\sqrt{\frac{y}{z+x}}$
$$+\sqrt{\frac{z}{x+y}}>2$$ | Proof: Using the binary mean inequality $2 \sqrt{a b} \leqslant a+b$. In fact,
$$\begin{array}{l}
\sqrt{\frac{x}{y+z}}+\sqrt{\frac{y}{z+x}}+\sqrt{\frac{z}{x+y}} \\
=\frac{2 x}{2 \sqrt{x(y+z)}}+\frac{2 y}{2 \sqrt{y(z+x)}}+\frac{2 z}{2 \sqrt{z(x+y)}} \\
\geqslant \frac{2 x}{x+(y+z)}+\frac{2 y}{y+(z+x)}+\frac{2 z}{z+(x+y)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,455 |
Example 9 Given that $a, b, c$ are positive numbers, $a+b+c=21$, prove:
$$a+\sqrt{a b}+\sqrt[3]{a b c} \leqslant 28$$ | Prove: By appropriate transformation and application of the two-variable and three-variable mean inequalities, we get
\[a+\sqrt{a b}+\sqrt[3]{a b c}=a+\sqrt{\frac{a}{2} \cdot 2 b}+\sqrt[3]{\frac{a}{4} \cdot b \cdot 4 c}\]
\[
\leqslant a+\frac{1}{2}\left(\frac{a}{2}+2 b\right)+\frac{1}{3}\left(\frac{a}{4}+b+4 c\right) \... | 28 | Inequalities | proof | Yes | Yes | inequalities | false | 736,456 |
Example 10 Let $a, b, c \in \mathbf{R}_{+}$, prove that for any real numbers $x, y, z$, we have
$$\begin{array}{l}
x^{2}+y^{2}+z^{2} \geqslant 2 \sqrt{\frac{a b c}{(a+b)(b+c)(c+a)}}\left(\sqrt{\frac{a+b}{c}} x y\right. \\
\left.+\sqrt{\frac{b+c}{a}} y z+\sqrt{\frac{c+a}{b}} z x\right)
\end{array}$$ | $$\begin{array}{l}
2 \sqrt{\frac{abc}{(a+b)(b+c)(c+a)}}\left(\sqrt{\frac{a+b}{c}} xy\right. \\
\left.+\sqrt{\frac{b+c}{a}} yz+\sqrt{\frac{c+a}{b}} zx\right) \\
=2 \sqrt{\frac{bx^2}{b+c} \cdot \frac{ay^2}{c+a}}+2 \sqrt{\frac{cy^2}{c+a} \cdot \frac{bz^2}{a+b}}+2 \sqrt{\frac{az^2}{a+b} \cdot \frac{cx^2}{b+c}} \\
\leqslant... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,457 |
Example 1 Given that $a, b, c$ are positive real numbers, $a^{2}+b^{2}+c^{2}+a b c$ $=4$, prove: $a+b+c \leqslant 3$. | Proof From the problem, we know that among $a, b, c$, there must be 2 that are not greater than 1, or not less than 1. Without loss of generality, let these be $b, c$. Thus, we have $(b-1)(c-1) \geqslant 0$,
which means $\square$
$$b c \geqslant b+c-1$$
By the two-variable mean inequality, we get
$$\begin{array}{c}
4-... | a+b+c \leqslant 3 | Inequalities | proof | Yes | Yes | inequalities | false | 736,458 |
Example 2 If real numbers $x, y, z$ satisfy $x y z=1$, prove: $x^{2}$
$$+y^{2}+z^{2}+3 \geqslant 2(x y+y z+z x) \text {. }$$ | Prove that since $x^{2} y^{2} z^{2}=1$, there must exist 2 out of $x^{2}, y^{2}, z^{2}$ that are both not greater than 1 or both not less than 1. Without loss of generality, assume these are $x^{2}, y^{2}$. Then we have $\left(\frac{1}{x^{2}}-1\right)\left(\frac{1}{y^{2}}-1\right) \geqslant 0$.
Because $x^{2}+y^{2}+z^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,459 |
Example 3 If $a, b, c$ are positive numbers, prove:
$$\frac{b^{2}}{a^{2}}+\frac{c^{2}}{b^{2}}+\frac{a^{2}}{c^{2}}+3 \geqslant\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right) .$$ | Proof: Let $\frac{a}{b}=x, \frac{b}{c}=y, \frac{c}{a}=z$, then the original inequality is equivalent to, if positive numbers $x, y, z$ satisfy $x y z=1$, then the inequality $\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+3 \geqslant 2(x+y+z)$ holds, which is $x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+3 \geqslant 2(x+y+z)$.... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,460 |
Example 4 For any positive real numbers $a, b, c$, we have
$$\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geqslant 3(a+b+c)^{2} .$$ | Prove that the original inequality can be transformed into
$$a^{2} b^{2} c^{2}+2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+\left(a^{2}+b^{2}+c^{2}\right)$$
$$+8 \geqslant 6(a b+b c+c a)$$.
By the Pigeonhole Principle, among \(a^{2}, b^{2}, c^{2}\), there must be two that are both not greater than 1 or both not le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,461 |
$$\begin{array}{l}
\text { Example } 5 \text { Let } a, b, c>0 \text {, prove that: } \\
\left(a^{5}-a^{3}+3\right)\left(b^{5}-b^{3}+3\right)\left(c^{5}-c^{3}+3\right) \geqslant 3(a \\
+b+c)^{2} \text {. }
\end{array}$$ | Prove that since $\left(a^{5}-a^{3}+3\right)-\left(a^{2}+2\right)=(a-1)^{2}(a+1)\left(a^{2}+a+1\right) \geqslant 0$, it follows that $a^{5}-a^{3}+3 \geqslant a^{2}+2$.
Similarly, we obtain two more inequalities. Multiplying the three inequalities, we get
$$\begin{array}{l}
\quad\left(a^{5}-a^{3}+3\right)\left(b^{5}-b^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,462 |
Example $\mathbf{6}$ Given $x, y, z \in \mathbf{R}_{+}$, prove:
$$\sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}} \leqslant \frac{3 \sqrt{2}}{2} .$$ | Proof: According to the Pigeonhole Principle, among $\sqrt{\frac{x}{x+y}}, \sqrt{\frac{y}{y+z}}, \sqrt{\frac{z}{z+x}}$, at least two are simultaneously not greater than the constant $\frac{\sqrt{2}}{2}$, or not less than the constant $\frac{\sqrt{2}}{2}$. Therefore, without loss of generality, let's assume $\sqrt{\frac... | \sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}} \leqslant \frac{3 \sqrt{2}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,463 |
Original problem: Given that $x, y$ are positive real numbers, $n \geqslant 2$ and $n \in \mathbf{N}$, prove or disprove:
$$\sqrt[n]{\frac{x}{\left(2^{n}-1\right) x+y}}+\sqrt[n]{\frac{y}{\left(2^{n}-1\right) y+x}} \leqslant 1$$ | Proof: First, prove the inequality: $\frac{x}{(t-1) x+y}+\frac{y}{(t-1) y+x} \leqslant \frac{2}{t} .(x, y$ are positive real numbers, $t \geqslant 2$ )
$$\begin{array}{l}
t x \cdot[(t-1) y+x]+t y \cdot[(t-1) x+y]-2[(t-1) x+y] \cdot[(t-1) y+x] \\
=t x^{2}+t y^{2}+2 t(t-1) x y-2\left[(t-1)^{2}+1\right] \cdot x y-2(t-1) x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,464 |
Theorem Let $M$ be a point inside a convex $n$-sided polygon $A_{1} A_{2} \cdots A_{n}(n \geqslant 3)$, and let $M B_{1}, M B_{2}, \cdots, M B_{n}$ be perpendicular to $A_{1} A_{2}, A_{2} A_{3}, \cdots, A_{n} A_{1}$, with the feet of the perpendiculars being $B_{1}, B_{2}, \cdots, B_{n}$, respectively. Then,
$$\begin{a... | Prove: Let $\angle M A_{i} A_{i+1}=\alpha_{i}\left(i=1, \cdots, n, A_{n+1}\right.$ $\equiv A_{1}$ ), then
$$\begin{array}{l}
P(M)=\prod_{i=1}^{n} \frac{M B_{i}}{M A_{i}}=\sin \alpha_{1} \cdots \sin \alpha_{n} . \\
\text { Also } P(M)=\prod_{i=1}^{n} \frac{M B_{i}}{M A_{i+1}}=\sin \left(A_{2}-\alpha_{2}\right) \cdots \s... | \left(\sin \frac{(n-2) \pi}{2 n}\right)^{n} | Geometry | proof | Yes | Yes | inequalities | false | 736,467 |
Question: Let $a, b \in(0,1)$, prove: $\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}} \geqslant \frac{a+b}{1-a b}$ $+\frac{a+b}{1-a b}\left(\frac{a-b}{1+a b}\right)^{2}$ | $\begin{array}{l}\text { Prove: } \frac{a}{1-a^{2}}+\frac{b}{1-b^{2}} \geqslant \frac{a+b}{1-a b}+\frac{a+b}{1-a b}\left(\frac{a-b}{1+a b}\right) \\ \Leftrightarrow \frac{a}{1-a^{2}}+\frac{b}{1-b^{2}} \geqslant \frac{(a+b)(1+a b)^{2}}{(1-a b)(1+a b)^{2}}+ \\ \frac{(a+b)(a-b)^{2}}{(1-a b)(1+a b)^{2}} \\ \Leftrightarrow ... | a^{4}+b^{4} \geqslant 2 a^{2} b^{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,472 |
Promotion: Let $a, b \in(0,1), n \in \mathbf{R}$, prove: $\frac{a^{n}}{1-a^{2 n}}+\frac{b^{n}}{1-b^{2 n}}$ $\geqslant \frac{a^{n}+b^{n}}{1-a^{n} b^{n}}+\frac{a^{n}+b^{n}}{1-a^{n} b^{n}}\left(\frac{a^{n}-b^{n}}{1+a^{n} b^{n}}\right)^{2}$ | $$\begin{array}{l}
\quad \text { Prove: } \frac{a^{n}}{1-a^{2 n}}+\frac{b^{n}}{1-b^{2 n}} \geqslant \frac{a^{n}+b^{n}}{1-a^{n} b^{n}}+\frac{a^{n}+b^{n}}{1-a^{n} b^{n}} \\
\left(\frac{a^{n}-b^{n}}{1+a^{n} b^{n}}\right)^{2} \\
\Leftrightarrow \frac{a^{n}-a^{n} b^{2 n}+b^{n}-a^{2 n} b^{n}}{\left(1-a^{2 n}\right)\left(1-b^... | a^{4 n}+b^{4 n} \geqslant 2 a^{2 n} b^{2 n} | Inequalities | proof | Yes | Yes | inequalities | false | 736,473 |
In $\triangle A B C$, the sides opposite to angles $A, B, C$ are $a, b, c$ respectively. If $A+C \leqslant 2 B$, prove:
1) For $1 \leqslant n \leqslant 4(n \in \mathbf{N})$, the inequality $a^{n}+$ $c^{n} \leqslant 2 b^{n}$ always holds;
2) For $n>4(n \in \mathrm{N})$, the inequality $a^{n}+c^{n} \leqslant$ $2 b^{n}$ d... | 1) From the original inequality $a^{4}+c^{4} \leqslant 2 b^{4}$, it is easy to see that equality holds if and only if $\cos B=\frac{1}{2}$, and $b^{2}=a c$, i.e., $a=b=c$. Therefore, we have
$$\left(a^{2}+c^{2}\right)^{2} \leqslant 2\left(a^{4}+c^{4}\right) \leqslant 2 \cdot 2 b^{4},$$
which implies
$$a^{2}+c^{2} \leq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,474 |
In $\triangle A B C$, the sides opposite to angles $A, B, C$ are $a, b, c$ respectively. If $A+C \leqslant 2 B$, when $n=\frac{1}{2}$, $\frac{1}{3}, \frac{1}{4}$, we have $a^{n}+c^{n} \leqslant 2 b^{n}$ | Prove using the arithmetic-geometric mean inequality and the conclusion $a+c \leqslant 2 b$, we get
$$\begin{aligned}
& \sqrt[4]{\frac{a}{b}}+\sqrt[4]{\frac{c}{b}} \\
= & \sqrt[4]{\frac{a}{b} \cdot 1 \cdot 1 \cdot 1}+\sqrt[4]{\frac{c}{b} \cdot 1 \cdot 1 \cdot 1} \\
\leqslant & \frac{1}{4}\left(\frac{a}{b}+1+1+1\right) ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,475 |
In $\triangle A B C$, the sides opposite to angles $A, B, C$ are $a, b, c$ respectively. If $A+C \leqslant 2 B$, then when $0<n \leqslant 4$,
$$a^{n}+c^{n} \leqslant 2 b^{n} \text {. }$$ | 1) First, prove the case where $n$ is a rational number.
Since the cases for $n=1,2,3,4$ have been proven in the previous text, it is only necessary to prove the case where $0 < n \leq 4$ and $q < 4p$. For any positive integer $q = 4u + i, i \in \{0,1,2,3\}$, we can write $q = 4u + 3v + 2x + y$, such that $v, x, y \in ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,476 |
Example 1 Given that $a$, $b$, $c$ are positive real numbers, $a^{2}+b^{2}+c^{2}+$ $abc=4$, prove: $a+b+c \leqslant 3$. | Proof: Among the positive real numbers $a$, $b$, $c$, there must be 2 that are simultaneously not less than 1, or not greater than 1. Without loss of generality, let these be $b$, $c$. Then we have $(b-1)(c-1) \geqslant 0$, which means $bc \geqslant b+c-1$.
Since $4-a^2 = b^2 + c^2 + abc \geqslant 2bc + abc = bc(2+a)$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,477 |
Example 1 (2004 National College Entrance Examination Chongqing Paper) Given the curve $y$ $=\frac{1}{3} x^{3}+\frac{4}{3}$, then the equation of the tangent line passing through point $P(2,4)$ is | It is known that point $P(2,4)$ lies on the curve $y=\frac{1}{3} x^{3}+\frac{4}{3}$, and $k_{P}=\left.y^{\prime}\right|_{x=2}=4$, so the equation of the tangent line is $y-$ $4=4(x-2)$, which simplifies to $4 x-y-4=0$. | 4 x-y-4=0 | Calculus | math-word-problem | Yes | Yes | inequalities | false | 736,480 |
Example 6 Given $a, b, c > 0$, prove:
$$\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geqslant 3(a+b+c)^{2} \text {. }$$ | Proof: Since $a^{2}-1, b^{2}-1, c^{2}-1$ must have 2 that are not greater than zero or not less than zero, let's assume they are $a^{2}-1, b^{2}-1$, thus we have $\left(a^{2}-1\right)\left(b^{2}-1\right) \geqslant 0$, which means $a^{2} b^{2}+1 \geqslant a^{2}+b^{2}$
which is $\left(a^{2}+2\right)\left(b^{2}+2\right) \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,481 |
Example 2 (Problem 159, Issue 3, 1988) In $\triangle A B C$, prove that: $\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}\left(\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}\right) \leqslant \frac{1}{3}$ | We know that in $\triangle A B C$, there is the identity
$$\begin{array}{l}
\tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2} \\
=1
\end{array}$$
Thus, this problem can be equivalently transformed into Problem 358 from Issue 2, 1995:
Let $x, y, z \in \mathbf{R}^{+},... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,486 |
It is said that in ancient times, Josephus and 31 other people were captured by barbarians, making them stand in a circle, numbered $1,2,3, \cdots, 32$, and then kill number 1, then kill number 3, and thereafter kill every other person, until the last person remaining was set free, and this person was Josephus himself.... | Solution: In the first round, kill $1,3,5, \cdots, 31$, leaving $2,4,6$, $\cdots, 32$; in the second round, kill $2,6,10, \cdots, 30$, leaving 4,8 , $12, \cdots, 32$; in the third round, kill $4,12,20,28$, leaving 8,16 , 24,32 ; in the fourth round, kill 8,24 , leaving 16,32 ; in the fifth round, kill 16 , leaving only... | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | inequalities | false | 736,492 |
Example 4 (Problem 508 of the 2nd Issue in 2000) Let $x, y, z \in \mathbf{R}^{+}$, prove that: $\sqrt{\frac{x}{y+z}}+\sqrt{\frac{y}{z+x}}+\sqrt{\frac{z}{y+x}}>2$. | This inequality (Macedonia, 1995) is quite elegant, and a concise proof can be given using the AM-GM inequality. In fact,
$$\begin{aligned}
& \sqrt{\frac{x}{y+z}}+\sqrt{\frac{y}{z+x}}+\sqrt{\frac{z}{x+y}} \\
= & \frac{2 x}{2 \sqrt{x(y+z)}}+\frac{2 y}{2 \sqrt{y(z+x)}}+\frac{2 z}{2 \sqrt{z(x+y)}} \\
\geqslant & \frac{2 x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,494 |
Example 5 (Problem 498, Issue 6, 1999) Given $0<x, y<1, n \in \mathrm{N}$, prove: $\frac{x^{n}}{1-x^{2}}+\frac{y^{n}}{1-y^{2}} \geqslant \frac{x^{n}+y^{n}}{1-x y}$ | This inequality is concise and clear. If the minus sign in the denominator is changed to a plus sign, it can give a similar inequality.
Given $x, y > 0, n \in \mathbf{N}^{*}$, prove:
$$\frac{x^{n}}{1+x^{2}}+\frac{y^{n}}{1+y^{2}} \leqslant \frac{x^{n}+y^{n}}{1+x y}.$$
Proof: By the two-variable mean inequality and the... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,495 |
Example 7 (Problem 728 of the 2nd Issue in 2008) Let the three sides of $\triangle ABC$ be $a, b, c$. Prove:
$$b \cos \frac{C}{2} + c \cos \frac{B}{2} < \frac{a + b + c}{\sqrt{2}}$$ | $$\begin{array}{l}
b \cos \frac{C}{2}+c \cos \frac{B}{2} \\
=\sqrt{b} \cdot \sqrt{b} \cos \frac{C}{2}+\sqrt{c} \cdot \sqrt{c} \cos \frac{B}{2} \\
\leqslant \sqrt{(b+c)\left(b \cos ^{2} \frac{C}{2}+c \cos ^{2} \frac{B}{2}\right)} \\
=\sqrt{(b+c)\left(b \cdot \frac{1+\cos C}{2}+c \cdot \frac{1+\cos B}{2}\right)} \\
=\sqr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,496 |
Example 8 (Problem 544, Issue 4, 2001) Given that $h_{a}, h_{b}, h_{c}, m_{a}, m_{b}, m_{c}$ are the altitudes and medians to the sides $a, b, c$ of $\triangle ABC$, respectively, prove:
$$\frac{m_{a}}{h_{b}+h_{c}}+\frac{m_{b}}{h_{c}+h_{a}}+\frac{m_{c}}{h_{a}+h_{b}} \geqslant \frac{3}{2}.$$
The original proof is quite... | Proof: On one hand, by the projection theorem, we get $a=b \cos C + c \cos B$. Therefore, applying the famous Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
a^{2}=(b \cos C+c \cos B)^{2} \\
\leqslant\left(b^{2}+c^{2}\right)\left(\cos ^{2} C+\cos ^{2} B\right)
\end{array}$$
Thus, $\cos ^{2} B+\cos ^{2} C \geqsla... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,497 |
Example 6 (Problem 553 from Issue 1, 2002) Let $x, y, z \in (0,1)$, prove that: $x(1-y)(1-z)+y(1-z)(1-x)+$ $z(1-x)(1-y)<1$ | The original proof constructs a linear function or constant function $F(x)$. In fact, from the condition, it is obvious that $0<1-x<1$, $0<1-y<1$, $0<1-z<1$. Therefore, the inequality can be strengthened to:
Given $x, y, z \in (0,1)$, prove: $x(1-y) + y(1-z) + z(1-x) < 1$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,498 |
Example 9 (Problem 289, Issue 6, 1992) Let $x, y, z$ be positive numbers. Prove that: $\frac{x y}{(y+z)(z+x)}+\frac{y z}{(z+x)(x+y)}+$ $\frac{z x}{(x+y)(y+z)} \geqslant \frac{3}{4}$ | Prove: Transforming the inequality into an integral inequality, we get
$$4 x y(x+y)+4 y z(y+z)+4 z x(z+x) \geqslant$$
$$3(x+y)(y+z)(z+x)$$, which is equivalent to $$x^{2} y+x y^{2}+y^{2} z+y z^{2}+z^{2} x+z x^{2} \geqslant 6 x y z$$.
Using the 6-variable mean inequality, the above inequality can be easily proven.
In ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,499 |
Let $a, b, c$ be positive numbers, $abc=1$, prove:
$$\begin{array}{l}
\left(a+\frac{1}{b}-1\right)\left(b+\frac{1}{c}-1\right) \\
+\left(b+\frac{1}{c}-1\right)\left(c+\frac{1}{a}-1\right) \\
+\left(c+\frac{1}{a}-1\right)\left(a+\frac{1}{b}-1\right) \geqslant 3 .
\end{array}$$ | Proof: According to the problem, let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}, x$, $y$, $z$ be positive numbers, then the inequality to be proved is equivalent to
$$\begin{array}{l}
\left(\frac{x}{y}+\frac{z}{y}-1\right)\left(\frac{y}{z}+\frac{x}{z}-1\right) \\
+\left(\frac{y}{z}+\frac{x}{z}-1\right)\left(\frac{z}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,500 |
Inverted 6 (1999 Shanghai College Entrance Exam Question) Find the set $A=|x||x-a|<2\}, B=$ $\left\{x \left\lvert\, \frac{2 x-1}{x+2}<1\right.\right\}$, if $A \subseteq B$, find the range of real number $a$.
---
The translation maintains the original text's formatting and structure, including the use of mathematical ... | Solution: From the given, we have $A=|x| a-2<x<a+2|$, $B=| x|-2<x<3|$. $\because A \subseteq B$, from the number line we get
$\therefore\left\{\begin{array}{l}a-2 \geqslant-2 \\ a+2 \leqslant 3\end{array}\right.$, thus $0 \leqslant a \leqslant 1$. | 0 \leqslant a \leqslant 1 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,501 |
Example 2 (Revised from the 1998 National College Entrance Examination) If $n$ is a natural number, prove:
$$(1+1)\left(1+\frac{1}{4}\right) \cdots\left(1+\frac{1}{3 n-2}\right)>\sqrt[3]{3 n+1}$$ | Prove: Construct a sequence, let
$$\begin{aligned}
\because \quad \frac{a_{n+1}}{a_{n}} & =\frac{\left(1+\frac{1}{3 n+1}\right) \cdot \sqrt[3]{3 n+1}}{\sqrt[3]{3 n+4}} \\
& =\sqrt[3]{\frac{(3 n+2)^{3}}{(3 n+4)(3 n+1)^{2}}} \\
& =\sqrt[3]{\frac{27 n^{3}+54 n^{2}+36 n+8}{27 n^{3}+54 n^{2}+27 n+4}} \\
& >1
\end{aligned}$$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,503 |
Example 3 (1985 Shanghai College Entrance Examination) For all natural numbers $n$ greater than 1, prove that
$$\left(1+\frac{1}{3}\right)\left(1+\frac{1}{5}\right) \cdots\left(1+\frac{1}{2 n-1}\right)>\frac{\sqrt{2 n+1}}{2}$$ | Prove: Construct the sequence $\left\{a_{n}\right\}$, let
$$\begin{array}{l}
a_{n}=\frac{2\left(1+\frac{1}{3}\right)\left(1+\frac{1}{5}\right) \cdots\left(1+\frac{1}{2 n-1}\right)}{\sqrt{2 n+1}} \text {, then } \\
\frac{a_{n-1}}{a_{n}}=\frac{\left(1+\frac{1}{2 n+1}\right) \cdot \sqrt{2 n+1}}{\sqrt{2 n+3}} \\
=\frac{2 n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,504 |
Example 4 (1992 "Three South" College Entrance Examination Question) Prove the inequality:
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}<2 \sqrt{n}(n \in \mathrm{N}) .$$ | Prove: Construct a sequence $a_{n}$, let
$$a_{n}=2 \sqrt{n}-\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}\right),$$
then $a_{n+1}-a_{n}=2 \sqrt{n+1}-2 \sqrt{n}-\frac{1}{\sqrt{n+1}}$
$$\begin{array}{l}
=\frac{2 n+1-2 \sqrt{n^{2}+n}}{\sqrt{n+1}} \\
=\frac{(\sqrt{n+1}-\sqrt{n})^{2}}{\sqrt{n+1}}>... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,505 |
Example 5 (1985 National College Entrance Examination Question) Let $a_{n}=\sqrt{1 \cdot 2}+\sqrt{2 \cdot 3}+\cdots+\sqrt{n(n+1)}(n=1$, $2, \cdots)$, prove the inequality: $\frac{n(n+1)}{2}<a_{n}<\frac{(n+1)^{2}}{2}$ for all positive integers $n$. | Prove: Construct the sequences $\left\{c_{n}\right\}, \left\{i b_{n}\right\}$, let
$c_{n}=a_{n}-\frac{(n+1)^{2}}{2}$, then
$$\begin{array}{l}
c_{n+1}-c_{n}=\sqrt{(n+1)(n+2)}-\frac{(n+2)^{2}}{2}+\frac{(n+1)^{2}}{2} \\
\quad=-\frac{(\sqrt{n+1})^{2}-2 \sqrt{(n+1)(n+2)}+(\sqrt{n+2})^{2}}{2} \\
=-\frac{(\sqrt{n+1}-\sqrt{n+2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,506 |
Example 1 Given that $x, y, z$ are positive real numbers, prove:
$$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \frac{3}{2} \text {. }$$ | Prove: Since $\frac{2 x}{y+z}+\frac{2 y}{z+x}+\frac{2 z}{x+y}-3$
$$\begin{aligned}
= & \left(\frac{2 x}{y+z}-1\right)+\left(\frac{2 y}{z+x}-1\right)+\left(\frac{2 z}{x+y}-1\right) \\
= & \frac{(x-y)-(z-x)}{y+z}+\frac{(y-z)-(x-y)}{z+x}+\frac{(z-x)-(y-z)}{x+y} \\
= & {\left[\frac{(x-y)}{y+z}-\frac{(x-y)}{z+x}\right]+\lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,507 |
Example 2 Let $a, b, c \in \mathbf{R}_{+}$, prove that:
$$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{a+b+c}{2} .$$ | Prove: By algebraic identity transformation, we get
$$\begin{aligned}
& \frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \\
= & \left(\frac{a^{2}}{b+c}+a\right)+\left(\frac{b^{2}}{c+a}+b\right)+\left(\frac{c^{2}}{a+b}+c\right)-(a+b+c) \\
= & \frac{a(a+b+c)}{b+c}+\frac{b(a+b+c)}{c+a}+\frac{c(a+b+c)}{a+b}-(a+b+c) \\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,508 |
Example 3 Given that $a, b, c$ are positive real numbers, prove:
$$\frac{a}{a b+a+1}+\frac{b}{b c+b+1}+\frac{c}{c a+c+1} \leqslant 1 .$$ | Proof: Using the method of difference comparison, and gradually finding a common denominator, we get
$$\begin{aligned}
& \frac{a}{a b+a+1}+\frac{b}{b c+b+1}+\frac{c}{c a+c+1}-1 \\
= & \frac{a}{a b+a+1}+\frac{b}{b c+b+1}-\frac{c a+1}{c a+c+1} \\
= & \left(\frac{a}{a b+a+1}-\frac{c a}{c a+c+1}\right)+\left(\frac{b}{b c+b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,509 |
Example 4 Let $\left|a_{i}\right|<1(i=1,2, \cdots, n)$, and denote $A_{1}(n)=a_{1}$ $+a_{2}+\cdots+a_{n}, A_{2}(n)=a_{1} a_{2}+a_{1} a_{3}+\cdots+a_{n-1} a_{n}, \cdots$, $A_{n}(n)=a_{1} a_{2} \cdots a_{n}$, prove that:
$$\left|\frac{A_{1}(n)+A_{3}(n)+A_{5}(n)+\cdots}{1+A_{2}(n)+A_{4}(n)+\cdots}\right|<1 .$$ | Prove: Let $H=\frac{A_{1}(n)+A_{3}(n)+A_{5}(n)+\cdots}{1+A_{2}(n)+A_{4}(n)+\cdots}$, it is easy to prove the identity $H=$
$$\frac{\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right)-\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n}\right)}{\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,510 |
Example 1 Given that the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are both arithmetic sequences, $S_{n}$ and $T_{n}$ are their respective sums of the first $n$ terms, and $\frac{S_{n}}{T_{n}}=\frac{7 n+1}{n+3}$, find the value of $\frac{a_{2}+a_{5}+a_{17}+a_{22}}{b_{8}+b_{10}+b_{12}+b_{16}}$. | Analysis: Students generally can use the property of the general term of an arithmetic sequence: "If $m, n, k, l \in \mathbf{N}^{*}$ and $m+n=k+l$, then $a_{m}+a_{n}=a_{k}+a_{l}$" to simplify the expression $\frac{a_{2}+a_{5}+a_{17}+a_{22}}{b_{8}+b_{10}+b_{12}+b_{16}}$. However, students often fail to establish the rel... | \frac{31}{5} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,511 |
Example 2 In the positive term geometric sequence $\left\{a_{n}\right\}$, $a_{1}+a_{2}+\cdots+a_{4 n}$ $=A, a_{1} \cdot a_{2} \cdot \cdots \cdot a_{4 n}=B$, express $\frac{1}{a_{1}}+\frac{1}{a_{2}}$ $+\cdots+\frac{1}{a_{4 n}}$ in terms of $A$, $B$, and $n$. | Analysis: It is not difficult to find that when the sequence $\left\{a_{n}\right\}$ is a positive geometric sequence, the sequence $\left\{\frac{1}{a_{n}}\right\}$ is also a geometric sequence, and its first term and common ratio are respectively equal to the
translation of the source text into English, preserving the... | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,512 |
Given the sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=\frac{3}{5}, a_{n+1}=\frac{3 a_{n}}{2 a_{n}+1}, n=1,2, \cdots$.
(1) Find the general term formula for $\left\{a_{n}\right\}$;
(2) Prove: For any $x>0, a_{n} \geqslant \frac{1}{1+x}-\frac{1}{(1+x)^{2}}\left(\frac{2}{3^{n}}-x\right), n=1,2, \cdots$;
(3)... | To prove the inequality using derivative knowledge, it is sufficient to find its maximum value as $a_{n}$. This change in thinking transforms the problem of proving an inequality into the quest for the maximum value of a function, a transformation that is reasonable and logical. This approach demonstrates a deep, broad... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,513 |
Question 2 Given the function
$$f_{t}(x)=\frac{1}{1+x}-\frac{1}{(1+x)^{2}}(t-x)$$
where the constant $t$ is a positive number.
(1) Find the maximum value of the function $f_{t}(x)$ on $(0,+\infty)$;
(2) In the sequence $\left\{a_{n}\right\}$, $a_{1}=3, a_{2}=5$, and its first $n$ terms sum $S_{n}$ satisfies $S_{n}+S_{... | Solution: (1) From $f_{t}(x)=\frac{1}{1+x}-\frac{1}{(1+x)^{2}}(t-x)$, we get $f_{t}^{\prime}(x)=-\frac{1}{(1+x)^{2}}-\frac{-(1+x)^{2}-2(t-x)(1+x)}{(1+x)^{4}}$ $=\frac{2(t-x)}{(1+x)^{3}}$.
Noting that $x>0$, so when $x0$; when $x>t$, $f_{t}^{\prime}(x)0$.
Let $t=\frac{1}{2^{n}}>0$,
then $f_{\frac{1}{2^{n}}}(x)=\frac{1}... | b_{1}+b_{2}+\cdots+b_{n}>\frac{n^{2}}{n+1} | Algebra | proof | Yes | Yes | inequalities | false | 736,514 |
Question 3 Let $x>0, y>0$.
(1) Prove: $\frac{1}{1+x} \geqslant \frac{1}{1+y}-\frac{1}{(1+y)^{2}}(x-y)$;
(2) Prove: $C_{n}^{0} \frac{3^{0}}{3^{0}+1}+C_{n}^{1} \frac{3^{1}}{3^{1}+1}+C_{n}^{2} \frac{3^{2}}{3^{2}+1}$ $+\cdots+C_{n}^{n} \frac{3^{n}}{3^{n}+1} \geqslant \frac{3^{n} \cdot 2^{n}}{3^{n}+2^{n}}$ | Prove: (1) The original inequality is equivalent to
$$\frac{1}{1+x} \geqslant \frac{1}{1+y}-\frac{1}{(1+y)^{2}}[(1+x)-(1+y)]$$
(Adding 1 and subtracting 1 is to transform the denominator structure)
$$=\frac{1}{1+y}-\frac{1+x}{(1+y)^{2}}+\frac{1}{1+y}$$
(The idea of combining and separating is well reflected here)
$$=\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,515 |
Question 1 Given $a b c=1$, prove: $\frac{a}{a b+a+1}+\frac{b}{b c+b+1}+\frac{c}{c a+c+1}=1$. | Prove that given $abc=1$, we can let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, then
$$\begin{aligned}
\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}= & \frac{\frac{x}{y}}{\frac{x}{y} \cdot \frac{y}{z}+\frac{x}{y}+1}+\frac{\frac{y}{z}}{\frac{y}{z} \cdot \frac{z}{x}+\frac{y}{z}+1}+\frac{\frac{z}{x}}{\frac{z}{x} ... | 1 | Algebra | proof | Yes | Yes | inequalities | false | 736,517 |
Question 3 Given real numbers $a, b, c$ such that the following fraction is defined, prove:
$$\frac{a}{a b+a+1}+\frac{b}{b c+b+1}+\frac{c}{c a+c+1}+\frac{(1-a b c)^{2}}{(a b+a+1)(b c+b+1)(c a+c+1)}=1$$ | $$\begin{array}{l}
\frac{a}{a b+a+1}+\frac{b}{b c+b+1}+\frac{c}{c a+c+1}-1=\frac{a}{a b+a+1}+\frac{b}{b c+b+1}-\frac{c a+1}{c a+c+1}= \\
\left(\frac{a}{a b+a+1}-\frac{c a}{c a+c+1}\right)+\left(\frac{b}{b c+b+1}-\frac{1}{c a+c+1}\right)= \\
\frac{a(1-a b c)}{(a b+a+1)(c a+c+1)}+\frac{a b c-1}{(b c+b+1)(c a+c+1)}= \\
\f... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,518 |
Example 1 Given $a, b, c \in \mathrm{R}^{+}$, prove: $\frac{a b}{a^{2}+a b+b c}+\frac{b c}{b^{2}+b c+c a}+\frac{c a}{c^{2}+c a+a b} \leqslant 1$. | Prove that, without loss of generality, let $abc=1$. First, we prove that
$$\frac{ab}{a^2+ab+bc} \leqslant \frac{b}{bc+b+1},$$
which is equivalent to
$$ab(bc+b+1) \leqslant b(a^2+ab+bc).$$
This simplifies to $abc + a \leqslant a^2 + bc$, and can be further reduced to $a + a^2 \leqslant a^3 + 1$. Since
$$(a^3 + 1) - (a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,521 |
Example 2 Given $x, y, z \in \mathrm{R}^{+}$, and $x y z=1$, prove: $\frac{1}{1+x+y}+\frac{1}{1+y+z}+\frac{1}{1+z+x} \leqslant 1$. | Prove that given the condition $x y z=1$, let $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$, and $a, b, c \in \mathrm{R}^{+}$, the inequality to be proven is equivalent to
$$\frac{a b}{a^{2}+a b+b c}+\frac{b c}{b^{2}+b c+c a}+\frac{c a}{c^{2}+c a+a b} \leqslant 1$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,522 |
Example 3 Let $a, b, c$ be positive numbers, and satisfy $a b c=1$, prove:
$$\frac{2}{(a+1)^{2}+b^{2}+1}+\frac{2}{(b+1)^{2}+c^{2}+1}+\frac{2}{(c+1)^{2}+a^{2}+1} \leqslant 1$$ | $$\begin{array}{l}
\frac{2}{(a+1)^{2}+b^{2}+1}+\frac{2}{(b+1)^{2}+c^{2}+1}+\frac{2}{(c+1)^{2}+a^{2}+1}= \frac{2}{a^{2}+b^{2}+2 a+2}+\frac{2}{b^{2}+c^{2}+2 b+2}+\frac{2}{c^{2}+a^{2}+2 c+2} \leqslant \\
\frac{2}{2 a b+2 a+2}+\frac{2}{2 b c+2 b+2}+\frac{2}{2 c a+2 c+2}= \\
\frac{1}{a b+a+1}+\frac{1}{b c+b+1}+\frac{1}{c a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,523 |
Example 1 If $x, y, z \in \mathrm{R}^{-}, x, z \geqslant 1$, prove:
$$\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z} \geqslant 2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) .$$ | Analysis: Transform the left expression and then apply the 3-variable inequality. In fact,
$$\begin{aligned}
\sum \frac{y+z}{x} & =\sum x \cdot \sum \frac{1}{x}-3 \\
& =\frac{1}{3} \sum x \cdot \sum \frac{1}{x}-3+\frac{2}{3} \sum-\frac{1}{x} \cdot \sum x \\
& \geqslant 3-3+\frac{2}{3} \sum-\frac{1}{x} \cdot 3 \sqrt[3]{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,525 |
Example 3 Let the three sides of $\triangle A B C$ be $a, b, c$, prove that
$$\begin{aligned}
& \frac{a}{\sqrt{2 b^{2}+2 a^{2}-a^{2}}}+\frac{b}{\sqrt{2 c^{2}+2 a^{2}-b^{2}}} \\
+ & \frac{c}{\sqrt{2 a^{2}+2 b^{2}-c^{2}}} \geqslant \sqrt{3} .
\end{aligned}$$ | Analysis: By using local substitution and then applying the 2-variable mean inequality to prove it.
Let $x=\frac{a}{\sqrt{2 b^{2}+2 c^{2}-a^{2}}}, y=\frac{b}{\sqrt{2 c^{2}+2 a^{2}-b^{2}}}, z=\frac{\varepsilon}{\sqrt{2 a^{2}+2 b^{2}-c^{2}}}$
Then $x^{2}=\frac{a^{2}}{2 b^{2}+2 c^{2}-a^{2}}$,
i.e., $\frac{x^{2}}{1+3 x^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,535 |
Example 5 Let $0<\theta<\pi$, find the maximum value of $y=\sin \frac{\theta}{2}\langle 1+\cos \theta\rangle$.
untranslated part:
$\langle 1+\cos \theta\rangle$
Note: The notation $\langle 1+\cos \theta\rangle$ is not standard and might be a typo or specific to the context. If it is meant to be parentheses, it sh... | Analysis: Eliminate the trigonometric function through substitution.
Let $\sin \frac{\theta}{2}=x, x \in(0,1)$, then
$$y=2 \sin \frac{\theta}{2}\left(1-\sin ^{2} \frac{\theta}{2}\right)=2 x\left(1-x^{2}\right)$$
Transforming, we get
$$\begin{aligned}
1 & =x^{2}+\frac{y}{2 x} \\
& =x^{2}+\frac{y}{4 x}+\frac{y}{4 x} \\
... | \frac{4 \sqrt{3}}{9} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,537 |
Example 6 Given $x, y, z \in \mathrm{R}^{-}, x+y+z=1$, find the minimum value of $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Example 6 Given $x, y, z \in \mathrm{R}^{-}, x+y+z=1$, f... | $$\begin{array}{l}
\text { Analysis: } \because x, y, z \in \mathbb{R}^{+}, x+y+z=1, \\
\therefore \quad \frac{1}{x}+\frac{4}{y}+\frac{9}{z} \\
=(x+y+z)\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right) \\
=\frac{2+z}{x}+\frac{4 z+4 x}{y}+\frac{9 x+9 y}{z}+14 \\
=\left(\frac{y}{x}+\frac{4 x}{y}\right)+\left(\frac{4 z}{y}... | 36 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,538 |
Let real numbers $a, b, c$ satisfy
$$\begin{array}{l}
a^{2}-b c-8 a+7=0 \\
b^{2}+c^{2}+h-6 a+6=0
\end{array}$$
Find the range of real number $a$. | $$\begin{array}{l}
1 b^{2}+c^{2}=-a^{2}+14 a-13, \\
12 b c=2 a^{2}-16 a+14 . \\
\because \quad b^{2}+c^{2} \geqslant 2 b c, \\
\therefore \quad-a^{2}+14 a-13 \geqslant 2 a^{2}-16 a+14,
\end{array}$$
That is, $a^{2}-10 a+9 \leqslant 0$.
Therefore, $1 \leqslant a \leqslant 9$. | 1 \leqslant a \leqslant 9 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,539 |
Example 9 Find all integer pairs that satisfy $\left(x^{2}+y^{2} \backslash x+y-3\right)=2 x y$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $$\begin{array}{l}
: \left.\begin{array}{l}
x=y, \\
x+y-3=1 \mathrm{i}
\end{array} \right\rvert\, \begin{array}{l}
x=-y, \\
x+y-3 \approx-1
\end{array} \\
\left\{\left.\begin{array}{l}
x y=0, \\
x+y-3=0 ;
\end{array} \right\rvert\, \begin{array}{l}
r=(1, \\
y=0 .
\end{array}\right.
\end{array}$$
Therefore, there are f... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,541 |
Question 2 Given that $x, y, z, r$ are all positive numbers, and $z \sqrt{x^{2}-r^{2}}=x^{2}, x y=r z$, prove: $x^{2}+y^{2}=z^{2}$.
| Analysis: There is the letter $r$ in the given conditions, but not in the conclusion, so, we just need to find a way to eliminate the letter (r).
For $z \sqrt{x^{2}-r^{2}}=x^{2}$, square both sides, we get $z^{2} x^{2}-$ $z^{2} r^{2}=x^{4}$
Substitute the condition $r z=x y$ into the above equation, to eliminate the ... | x^{2}+y^{2}=z^{2} | Algebra | proof | Yes | Yes | inequalities | false | 736,543 |
Question 3 Given that $x, y, z, r$ are all positive numbers, and $x^{2}+y^{2}=z^{2}, x y=r z$, prove: $z \sqrt{x^{2}-r^{2}}=x^{2}$. | Analysis: There is a letter $y$ in the conditional relationship, but there is none in the conclusion, so, combine the two equations given in the condition to eliminate the letter $y$.
From $x y=r z$, solve for $y=\frac{r z}{x}$, substitute into $x^{2}+y^{2}=z^{2}$, and perform targeted transformations with the goal in... | z \sqrt{x^{2}-r^{2}}=x^{2} | Algebra | proof | Yes | Yes | inequalities | false | 736,544 |
Question $1-2$ Given that $x, y, z, r$ are all positive numbers, and $z \sqrt{x^{2}-r^{2}}=x^{2}, x y=r z$, prove: $x^{2}+y^{2}=z^{2}$. | Analysis: The letter $r$ appears in the given conditions, but not in the conclusion, so the goal is to eliminate the letter $r$.
For $z \sqrt{x^{2}-r^{2}}=x^{2}$, square both sides to get $z^{2} x^{2}-z^{2} r^{2}=x^{4}$.
Substitute the condition $r z=x y$ into the above equation to eliminate the letter $r$, resulting ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,546 |
Given $a b c d=1$, prove:
$$\begin{aligned}
& \frac{a}{a b c+a b+a+1}+\frac{b}{b c d+b c+b+1}+\frac{c}{c d a+c d+c+1} \\
+ & \frac{d}{d a b+d a+d+1}=1
\end{aligned}$$ | Analysis: Eliminate the letter $d$. For the fractions on the left side of the equation to be proven, starting from the 2nd, multiply the numerator and denominator by $a, a b, a b c$ respectively, to get
$$\begin{array}{l}
\frac{a}{a b c+a b+a+1}+\frac{b}{b c d+b c+b+1}+\frac{c}{c d a+c d+c+1} \\
+\frac{d}{d a b+d a+d+1... | 1 | Algebra | proof | Yes | Yes | inequalities | false | 736,548 |
Promotion 1 Given $a>0, b>0, ab=1$, prove:
$$\frac{b}{1+a^{4}}+\frac{a}{1+b^{4}} \geqslant 1$$ | $\begin{array}{l}\text { Prove } \quad \frac{b}{1+a^{4}}+\frac{a}{1+b^{4}}=\frac{b^{3}}{b^{2}+a^{4} b^{2}}+\frac{a^{3}}{a^{2}+a^{2} b^{4}} \\ =\frac{a^{3}+b^{3}}{a^{2}+b^{2}} \geqslant \frac{a+b}{2} \geqslant \sqrt{a b}=1 .\end{array}$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,550 |
Promotion 2 Given $a, b \in \mathbf{R}^{+}, ab=1$, prove:
$$\frac{b}{1+a^{3}}+\frac{a}{1+b^{3}} \geqslant 1$$ | Prove that
$$\begin{array}{l}
\frac{b}{1+a^{3}}+\frac{a}{1+b^{3}}=\frac{a b}{a\left(1+a^{3}\right)}+\frac{a}{1+a^{-3}} \\
=\frac{1+a^{5}}{a\left(1+a^{3}\right)}
\end{array}$$
Therefore, we only need to prove that $\frac{1+a^{5}}{a\left(1+a^{3}\right)} \geqslant 1$, which is equivalent to $1+a^{5} \geqslant a+a^{4}$. I... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,551 |
Promotion 3 Given $a>0, b>0, ab=1$, prove: $a^{3}+b^{3}$ $\geqslant a+b$ | Prove by using the method of difference, factorization, we get
$$\begin{array}{l}
\left(a^{3}+b^{3}\right)-(a+b) \\
=(a+b)\left(a^{2}+b^{2}-a b\right)-(a+b) a b \\
=(a+b)(a-b)^{2} \geqslant 0 . \\
\therefore a^{3}+b^{3} \geqslant a+b .
\end{array}$$
This inequality, when transformed into a one-variable case, yields | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,552 |
Example 1 Given positive numbers $x, y$ satisfying $x+y=1$, prove:
$$\sqrt{2 x^{2}+y}+\sqrt{x+2 y^{2}} \geqslant 2$$ | Analysis and Proof: To rationalize an irrational expression, the technique of squaring is required. The original inequality is equivalent to
$$2 x^{2}+y+x+2 y^{2}+2 \sqrt{\left(2 x^{2}+y\right)\left(x+2 y^{2}\right)} \geqslant 4 \text {. }$$
By grouping similar terms, we obtain a symmetric and elegant inequality: $2\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,553 |
Example 2 Let $a, b, c$ be positive real numbers, prove that:
$$\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geqslant \frac{3}{1+a b c} .$$ | Analysis and Proof: Eliminating the denominator on the right side of the inequality, we can multiply both sides of the original inequality by \(1 + abc\). Thus, the original inequality is equivalent to
$$\frac{1+abc}{a(1+b)}+\frac{1+abc}{b(1+c)}+\frac{1+abc}{c(1+a)} \geqslant 3$$
Next, add 3 to both sides of this ineq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,554 |
Example 3 Let $a, b, c$ be positive real numbers, prove that: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}$. | Analysis and Proof: By using local substitution, let $x=\frac{a}{b+c}$, $y=\frac{b}{c+a}$, $z=\frac{c}{a+b}\left(x, y, z \in \mathbf{R}_{+}\right)$. Next, we aim to eliminate the letters $a$, $b$, and $c$, and transform the problem into one involving the variables $x$, $y$, and $z$.
Adding 1 to both sides of $x=\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,555 |
Example 3 Given that $a, b, c$ are positive real numbers, and $a+b+c=12, ab+bc$ $+ca=45$, try to find the maximum value of $abc$.
保留了原文的换行和格式。 | Solve: From $a+b+c=12$, we get $a+c=12-b$, substitute into the transformed form of $ab+bc+ca=45$, which is $b(a+c)+ca=45$,
we get $b(12-b)+ca=45$, that is, $ca=b^{2}-12b+45$.
Thus, $b, c$ are the two roots of the following quadratic equation
$$t^{2}-(12-b)t+\left(b^{2}-12b+45\right)=0$$
Since $t$ is a real number, th... | 54 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,559 |
Example 4 Given $a b c=1$, prove: $\frac{a}{a b+a+1}+\frac{b}{b c+b+1}+$
保留源文本的换行和格式，直接输出翻译结果。 | $$\begin{array}{l}
\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1} \\
=\frac{\frac{x}{y}}{\frac{x}{y} \cdot \frac{y}{z}+\frac{x}{y}+1}+\frac{\frac{y}{z}}{\frac{y}{z} \cdot \frac{z}{x}+\frac{y}{z}+1} \\
+\frac{\frac{z}{x}}{\frac{z}{x} \cdot \frac{x}{y}+\frac{z}{x}+1} \\
=\frac{zx}{xy+zx+yz}+\frac{xy}{yz+xy+zx}+\frac{... | 1 | Algebra | proof | Yes | Yes | inequalities | false | 736,560 |
Example 1 (Shandong) Let the line $l$ with a slope of 2 pass through the focus $F$ of the parabola $y^{2}=$ $a x(a \neq 0)$, and intersect the $y$-axis at point $A$. If the area of $\triangle O A F$ (where $O$ is the origin) is 4, then the equation of the parabola is ( ).
A. $y^{2}= \pm 4 x$
B. $y^{2}= \pm 8 x$
C. $y^{... | Explain that for the parabola $y^{2}=a x(a \neq 0)$, the coordinates of the focus $F$ are $\left(\frac{a}{4}, 0\right)$. Therefore, the equation of the line $l$ is $y=2\left(x-\frac{a}{4}\right)$, and its intersection with the $y$-axis is $A\left(0,-\frac{a}{2}\right)$. Thus, the area of $\triangle O A F$ is $\frac{1}{... | B | Geometry | MCQ | Yes | Yes | inequalities | false | 736,562 |
Example 2 (National I) If $\frac{\pi}{4}<x<\frac{\pi}{2}$, then the maximum value of the function $y=$ $\tan 2x \tan ^{3} x$ is | Given $\frac{\pi}{4}1$, we have $y=$ $\frac{2 \tan ^{4} x}{1-\tan ^{2} x}$. Let $t=1-\tan ^{2} x$, then $t \in(-\infty, 0)$. At this point, transforming it into a fractional function of $t$, and using the arithmetic mean-geometric mean inequality, we get
$$y=\frac{2(1-t)^{2}}{t}=-\left(-2 t+\frac{2}{-t}\right)-4 \leqsl... | -8 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,563 |
Example 3 (National I) In the right triangular prism $A B C-A_{1} B_{1} C_{1}$, all vertices lie on the same sphere. If $A B=A C=A A_{1}=2, \angle B A C=$ $120^{\circ}$, then the surface area of this sphere is $\qquad$ | As shown in Figure 1, take the midpoint $M$ of $BC$, connect $AM$ and extend it to $D$ such that $AD=4$. It is easy to prove that $\triangle ABD$ and $\triangle ACD$ are both right triangles. Therefore, quadrilateral $ABDC$ can be inscribed in a circle. Thus, we can use quadrilateral $ABDC$ as the base to complete the ... | 20\pi | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,564 |
Example 4 (Beijing) Let $D$ be the set of points consisting of the equilateral $\triangle P_{1} P_{2} P_{3}$ and its interior, and let point $P_{0}$ be the center of $\triangle P_{1} P_{2} P_{3}$. If the set $S=$ $\left\{P|P \in D|, P P_{0}|\leqslant| P P_{i} \mid, i=1,2,3\right\}$, then the planar region represented b... | As shown in the figure, $A, B, C, D, E, F$ are the trisection points of each side. From $\left|P P_{0}\right| \leqslant\left|P P_{1}\right|$, we know that point $P$ is on the perpendicular bisector of segment $P_{0} P_{i}$ or on the side closer to point $P_{0}$. Since point $P \in D$, then $S$ represents the regular he... | D | Geometry | MCQ | Yes | Yes | inequalities | false | 736,565 |
Example 5 (Shaanxi) In a class, 36 students participate in mathematics, physics, and chemistry extracurricular research groups. Each student participates in at most two groups. It is known that the number of students participating in the mathematics, physics, and chemistry groups are 26, 15, and 13, respectively. The n... | From the problem, we know that a set model should be constructed. Let the sets representing the number of people participating in the mathematics, physics, and chemistry groups be $A, B, C$, respectively. Then, card $(A \cap B \cap C)=0, \operatorname{card}(A \cap B)=6, \operatorname{card}(B \cap C)=4$.
Using the form... | 8 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 736,566 |
Example 6 (Xi) Given the sequence $\left\{x_{n}\right\}$ satisfies $x_{1}=\frac{1}{2}, x_{n+1}=$ $\frac{1}{1+x_{n}}, n \in \mathbf{N}^{*}$.
(I) Conjecture the monotonicity of the sequence $\left\{x_{2 n}\right\}$, and prove your conclusion;
(II) Prove: $\left|x_{n+1}-x_{n}\right| \leqslant \frac{1}{6}\left(\frac{2}{5}\... | (I) Given $x_{1}=\frac{1}{2}, x_{n+1}=\frac{1}{1+x_{n}}$, we calculate $x_{2}=\frac{2}{3}, x_{3}=\frac{3}{5}, x_{4}=\frac{5}{8}, x_{5}=\frac{8}{13}, x_{6}=\frac{13}{21}$,
thus, we have $x_{2}>x_{4}>x_{6}$.
Therefore, we conjecture that the sequence $\left\{x_{2 n}\right\}$ is a decreasing sequence.
We will prove this b... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,567 |
Given $a>0, b>0, a+b=1$, prove that
$$\sqrt{2}<\sqrt{a+\frac{1}{2}}+\sqrt{b+\frac{1}{2}} \leqslant 2$$ | Prove $\because a+b=1$,
$\therefore\left(\sqrt{a+\frac{1}{2}}+\sqrt{b+\frac{1}{2}}\right)^{2}$
$=2+2 \sqrt{a b+\frac{3}{4}}$
and $a>0, b>0$,
from $1=a+b \geqslant 2 \sqrt{a b}$ we know $0<a b \leqslant \frac{1}{4}$.
$\therefore 2+\sqrt{3}<\left(\sqrt{a+\frac{1}{2}}+\sqrt{b+\frac{1}{2}}\right)^{2} \leqslant 4$.
Taking t... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,569 |
Example 1: If $a, b \in(0,1)$, then $\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}} \geq \frac{a+b}{1-a b}$ | Proof:
$$\begin{aligned}
\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}} & =\frac{(a+b)(1-a b)}{1-\left(a^{2}+b^{2}\right)+a^{2} b^{2}} \\
\geq & \frac{(a+b)(1-a b)}{1-2 a b+a^{2} b^{2}}=\frac{a+b}{1-a b} .
\end{aligned}$$
Similarly: If $a, b \in(0,1)$, then
(1) $\frac{a}{1-b^{2}}+\frac{b}{1-a^{2}} \geq \frac{a+b}{1-a b}$
(2) $\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,571 |
Example 10: Given that $x, y, z$ are positive real numbers, prove:
$$\frac{x}{2 x+y+z}+\frac{y}{x+2 y+z}+\frac{z}{x+y+2 z} \leq \frac{3}{4} .$$ | Prove that for $a, b \in \mathbb{R}^{+}$, it is obvious that
$$\frac{1}{a+b} \leq \frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)$$
Thus,
$$\begin{aligned}
& \frac{x}{2 x+y+z}+\frac{y}{x+2 y+z}+\frac{z}{x+y+2 z} \\
\leq & \frac{x}{4}\left(\frac{1}{z+x}+\frac{1}{x+y}\right)+\frac{y}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\... | \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 736,573 |
Example 11: (Problem 159, Issue 3, 1988, "Mathematics Teaching") In $\triangle ABC$, prove:
$$\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}\left(\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}\right) \leq \frac{1}{3}$$ | We know that, in $\triangle A B C$, there is the identity
$$\tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} = 1.$$
Thus, this problem is equivalent to Problem 358 in the 2nd issue of 1995:
Let $x, y, z \in \mathbb{R}^{+}, x y + y z + z x = 1$, prove that:
$$x y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,574 |
Example 12: Let the three sides of $\triangle A B C$ be $a, b, c$, prove:
$$2\left(\cos ^{2} A+\cos ^{2} B+\cos ^{2} C\right) \geq \frac{a^{2}}{b^{2}+c^{2}}+\frac{b^{2}}{c^{2}+a^{2}}+\frac{c^{2}}{a^{2}+b^{2}} \geq \frac{3}{2}$$ | Proof: On one hand, by the projection theorem, we get $a=b \cos C+c \cos B$. Then, applying the famous Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
a^{2}=(b \cos C+c \cos B)^{2} \leq\left(b^{2}+c^{2}\right)\left(\cos ^{2} C+\cos ^{2} B\right) \\
\cos ^{2} B+\cos ^{2} C \geq \frac{a^{2}}{b^{2}+c^{2}}
\end{array... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,575 |
Example 13: (Problem 544 from Issue 4, 2001 of "Mathematics Teaching") Given that $h_{a}, h_{b}, h_{c}, m_{a}, m_{b}, m_{c}$ are the lengths of the altitudes and angle bisectors on the sides $a, b, c$ of $\triangle A B C$, respectively, prove:
$$\frac{m_{a}}{h_{b}+h_{c}}+\frac{m_{b}}{h_{c}+h_{a}}+\frac{m_{c}}{h_{a}+h_{... | In $\triangle A B C$, the heights and angle bisectors on the three sides $a, b, c$ are $h_{a}, h_{b}, h_{c}$ and $w_{a}, w_{b}, w_{c}$, respectively. Prove:
$$\frac{w_{a}}{h_{b}+h_{c}}+\frac{w_{b}}{h_{c}+h_{a}}+\frac{w_{c}}{h_{a}+h_{b}} \geq \frac{3}{2}$$
Similarly, note that $w_{a} \geq h_{a}, w_{b} \geq h_{b}, w_{c}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,576 |
Example 14: (Problem 553 from Issue 1, 2002) Let $x, y, z \in(0,1)$, prove that:
$$x(1-y)(1-z)+y(1-z)(1-x)+z(1-x)(1-y)<1 .$$ | The original proof constructs a linear function or constant function $F(x)$. In fact, from the condition, it is obvious that $0<1-x<1$, $0<1-y<1$, $0<1-z<1$. Therefore, the inequality can be strengthened to:
Given $x, y, z \in(0,1)$, prove:
$$x(1-y)+y(1-z)+z(1-x)<1$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,577 |
Example 15: If real numbers $a, b, c$ satisfy $a^{2}+b^{2}+c^{2}+2 a b c=1$, prove:
$$\sqrt{\frac{a^{2}+b^{2}}{1+a^{2} b^{2}}}+\sqrt{\frac{b^{2}+c^{2}}{1+b^{2} c^{2}}}+\sqrt{\frac{c^{2}+a^{2}}{1+c^{2} a^{2}}} \leq 3 .$$ | To prove: By regarding the conditional equation as a quadratic equation in $c$, we get
$$c^{2}+2 a b \cdot c+\left(a^{2}+b^{2}-1\right)=0$$
Since $c \in \mathbb{R}$, the discriminant $\Delta=(2 a b)^{2}-4\left(a^{2}+b^{2}-1\right) \geq 0$,
which means $1+a^{2} b^{2} \geq a^{2}+b^{2}$, thus $\frac{a^{2}+b^{2}}{1+a^{2} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,578 |
Example 16: If $a, b, c \in\left[\frac{1}{\sqrt{2}}, \sqrt{2}\right]$, prove that: $\frac{a b+b c}{a^{2}+2 b^{2}+c^{2}} \geq \frac{2}{5}$. | Proof: It is easy to prove that the function $y=x+\frac{1}{x}$ is a decreasing function on $\left[\frac{1}{2}, 1\right]$ and an increasing function on $[1,2]$. Therefore, when $x=\frac{1}{2}$ or $x=2$, the maximum value of the function $y=x+\frac{1}{x}$ on $\left[\frac{1}{2}, 2\right]$ is $y_{\text {max }}=\frac{5}{2}$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,579 |
Example 17: Given $a, b, c \in R^{+}, a b+b c+c a=a b c$, prove:
$$\frac{a^{5}+b^{5}}{a b\left(a^{4}+b^{4}\right)}+\frac{b^{5}+c^{5}}{b c\left(b^{4}+c^{4}\right)}+\frac{c^{5}+a^{5}}{c a\left(c^{4}+a^{4}\right)} \geq 1$$ | Prove: Transform the inequality to be proved into:
$$\frac{a^{-5}+b^{-5}}{a^{-4}+b^{-4}}+\frac{b^{-5}+c^{-5}}{b^{-4}+c^{-4}}+\frac{c^{-5}+a^{-5}}{c^{-4}+a^{-4}} \geq 1$$
Let $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$, then we have $x, y, z \in R^{+}, x+y+z=1$, at this time, the inequality to be proved is equivalent... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,580 |
Example 18: If $a, b \in R_{+}, a+b=1$, prove:
(1) $\frac{a}{a^{2}+b}+\frac{b}{a+b^{2}} \leq \frac{4}{3}$;
(2) $\frac{a}{a+b^{3}}+\frac{b}{a^{3}+b} \leq \frac{8}{5}$. | Proof: (1) Let $t=ab$, then $t \in \left(0, \frac{1}{4}\right]$. From $a+b=1$, we get
$a^{2}+b^{2}=1-2ab, \quad a^{3}+b^{3}=1-3ab$. Therefore,
$$\begin{array}{l}
\frac{a}{a^{2}+b}+\frac{b}{a+b^{2}} \leq \frac{4}{3} \\
\Leftrightarrow 3a\left(a+b^{2}\right)+3b\left(a^{2}+b\right) \leq 4\left(a^{2}+b\right)\left(a+b^{2}\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,581 |
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