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Example 2 (2003 IMO USA National Team Selection Exam Question) Let $a, b, c$ be real numbers in the interval $\left(0, \frac{\pi}{2}\right)$. Prove:
$$\begin{array}{l}
\frac{\sin a \sin (a-b) \sin (a-c)}{\sin (b+c)}+\frac{\sin b \sin (b-c) \sin (b-a)}{\sin (c+a)} \\
+\frac{\sin c \sin (c-a) \sin (c-b)}{\sin (a+b)} \geq... | Prove: Using the product-to-sum formulas and the double-angle formulas, we have
$$\begin{array}{l}
\sin a \sin (a-b) \sin (a-c) \sin (a+b) \sin (a+c) \\
=\sin a\left(\sin ^{2} a-\sin ^{2} b\right)\left(\sin ^{2} a-\sin ^{2} c\right)
\end{array}$$
Let \( x = \sin^2 a, y = \sin^2 b, z = \sin^2 c \), then the inequality ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,274 |
Example 3 (2001 IMO Romanian National Team Selection Test) Given that $a, b, c$ are the lengths of the three sides of a triangle, prove:
$$\begin{array}{l}
\quad(-a+b+c)(a-b+c)+(a-b+c)(a+b-c) \\
+(a+b-c)(-a+b+c) \leqslant \sqrt{a b c}(\sqrt{a}+\sqrt{b}+\sqrt{c}) .
\end{array}$$ | Proof: Assume $a, b, c$ are any positive numbers. After standardization, the original inequality becomes
$$2(ab + bc + ca) \leq a^2 + b^2 + c^2 + a \sqrt{bc} + b \sqrt{ca} + c \sqrt{ab}.$$
Let $a = x^2, b = y^2, c = z^2$, then the above inequality is equivalent to
$$\begin{array}{l}
x^4 + y^4 + z^4 + x^2 y z + x y^2 z ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,275 |
Example 4 (Crux2006: $415-416$ ) Let $k>-1$ be a fixed real number, and $a, b, c$ be non-negative real numbers, satisfying $a+b+c=1$, and $ab + bc + ca > 0$. Find: $\min \left\{\frac{(1+ka)(1+kb)(1+kc)}{(1-a)(1-b)(1-c)}\right\}$. | Solution: The required minimum value is $\min \left\{\frac{1}{8}(k+3)^{3},(k+2)^{2}\right\}$. First, we prove the following inequality:
$4(a b+b c+c a) \leqslant 1+9 a b c$.
Using the condition $a+b+c=1$, we get
$$\begin{array}{l}
1-4(a b+b c+c a)+9 a b c \\
=(a+b+c)^{3}-4(a+b+c)(a b+b c+c a)+9 a b c \\
=a(a-b)(a-c)+b(... | \min \left\{\frac{1}{8}(k+3)^{3},(k+2)^{2}\right\} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,276 |
Example 5 (Crux2006:190-191) Let $a, b, c$ be non-negative real numbers, satisfying $a^{2}+b^{2}+c^{2}=1$, prove:
$$\frac{1}{1-a b}+\frac{1}{1-b c}+\frac{1}{1-c a} \leqslant \frac{9}{2} .$$ | Prove: The original inequality is equivalent to
$$\begin{array}{l}
3-5(a b+b c+c a)+6 a b c(a+b+c)+a b c(a+b \\
+c-9 a b c) \geqslant 0 .
\end{array}$$
By the AM-GM inequality, we have
$$\begin{array}{l}
a+b+c-9 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)-9 a b c \\
\geqslant 3 \sqrt[3]{a b c} \cdot 3 \sqrt[3]{a^{2} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,277 |
Example 6 (1996 Iran) Prove: For positive real numbers $x, y, z$, the following inequality holds
$$\begin{aligned}
& (x y+y z+z x)\left[\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}}\right] \\
\geqslant & \frac{9}{4} .
\end{aligned}$$ | $$\begin{array}{l}
\text { Prove: By clearing the denominators, the original inequality becomes } \\
\quad \sum_{s y m}\left(4 x^{5} y-x^{4} y^{2}-3 x^{3} y^{3}+x^{4} y z-2 x^{3} y^{2} z\right. \\
\left.+x^{2} y^{2} z^{2}\right) \geqslant 0,
\end{array}$$
Here $\sum_{s y m}$ denotes the sum over all permutations of $x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,278 |
Example 7 (1997 Japan) Let $a, b, c$ be positive real numbers. Prove that:
$$\frac{(b+c-a)^{2}}{(b+c)^{2}+a^{2}}+\frac{(c+a-b)^{2}}{(c+a)^{2}+b^{2}}+\frac{(a+b-c)^{2}}{(a+b)^{2}+c^{2}} \geqslant \frac{3}{5}$$
and determine the conditions under which equality holds. | Prove: Simplify the original inequality, we get
$$\sum_{s m m} \frac{2 a b+2 a c}{a^{2}+b^{2}+c^{2}+2 b c} \leqslant \frac{12}{5} .$$
Let $s=a^{2}+b^{2}+c^{2}$, by clearing the denominators, we get
$$5 s^{2} \sum_{s m} a b+10 s \sum_{s, m} a^{2} b c+20 \sum_{s y m} a^{3} b^{2} c$$
$$\leqslant 6 s^{3}+6 s^{2} \sum_{v m... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,279 |
Example 8 Let $a \geqslant b \geqslant c$, non-negative real numbers $x, y, z$ satisfy $x+z \geqslant y$. Prove: $x^{2}(a-b)(a-c)+y^{2}(b-c)(b-a)+z^{2}(c-a)(c-b) \geqslant 0$. | \begin{aligned} & \text { Proof: Since } x+z \geqslant y, \text { and }(b-c)(b-a) \leqslant 0, \\ & \quad \text { then } x^{2}(a-b)(a-c)+y^{2}(b-c)(b-a)+z^{2}(c \\ -a) & (c-b) \\ & \geqslant x^{2}(a-b)(a-c)+(x+z)^{2}(b-c)(b-a) \\ +z^{2} & (c-a)(c-b) \\ & =[x(a-b)+z(c-b)]^{2} \geqslant 0 . \\ & \text { The equality hold... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,280 |
Example 9 Given real numbers $a, b, c, x, y, z$ satisfy $a \geqslant b \geqslant c$, and $x \geqslant y \geqslant z$ or $x \leqslant y \leqslant z, k$ is a positive integer. $f: \mathbf{R} \rightarrow \mathbf{R}_{+}$ is a monotonic or convex function, prove:
$$\begin{array}{l}
f(x)(a-b)^{k}(a-c)^{k}+f(y)(b-c)^{k}(b-a)... | Prove: When $k$ is even, the inequality obviously holds.
When $k$ is odd, since $f: \mathbf{R} \rightarrow \mathbf{R}_{+}$ is a monotonic or convex function, and $x \geqslant y \geqslant z$ or $x \leqslant y \leqslant z$, then $f(x) \geqslant f(y)$ and $f(z) \geqslant f(y)$ must have one that holds, without loss of gen... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,281 |
Inference 1 The necessary and sufficient condition for the symmetric cubic inequality satisfying $P_{3}(1,1,1)=0$
$$P_{3}(x, y, z)=\lambda_{0,1} \sigma_{1}^{3}+\lambda_{0,2} \sigma_{1} \sigma_{2}+\lambda_{1,1} \sigma_{3} \geq 0$$
is
$$P_{3}(1,0,0)=\lambda_{0.1} \geq 0, \quad P_{3}(1,1,0)=2\left(4 \lambda_{0,1}+\lambda... | Proof: The necessity is obviously true. Below we prove the sufficiency.
Since $P_{3}(1,1,1)=0$, according to the algorithm Schp, we can obtain
$$P_{3}(x, y, z)=\alpha_{0.1} f_{0.1}^{(3)}+\alpha_{0.2} f_{0,2}^{(3)}=\lambda_{0.1} f_{0.1}^{(3)}+\left(4 \lambda_{0.1}+\lambda_{0,2}\right) f_{0.2}^{(3)},$$
Therefore, if equ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,283 |
Inference 3 Symmetric Inequality of Degree 6
$$P_{6}(x, y, z)=\lambda_{0,3} \sigma_{1}^{2} \sigma_{2}^{2}+\lambda_{0,4} \sigma_{2}^{3}+\sigma_{3}\left(\lambda_{1,1} \sigma_{1}^{3}+\lambda_{1,2} \sigma_{1} \sigma_{2}\right)+\lambda_{2,1} \sigma_{3}^{2} \geq 0$$
Under the condition $P_{6}(1,1,1)=0$, the necessary and su... | Proof According to the method Schp, we can obtain
$$P_{6}=\alpha_{0.3} f_{0,3}^{(6)}+\alpha_{0.4} f_{0.4}^{(6)}+\alpha_{1,1} f_{1,1}^{(6)}+\alpha_{1.2} f_{1,2}^{(6)}$$
where $\alpha_{0,3}=\lambda_{0,3}, \alpha_{0,4}=4 \lambda_{0,3}+\lambda_{0,4}, \alpha_{1,1}=4 \lambda_{0,3}+\lambda_{1,1}, \alpha_{1,2}=14 \lambda_{0,3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,284 |
Example 1 Prove
$$S \leq \frac{\sqrt{3} \sqrt[3]{a b c} \sqrt[6]{a^{2} b^{2} c^{2}-(a-b)^{2}(b-c)^{2}(c-a)^{2}}}{4}$$
where $a, b, c, S$ are the three sides and the area of an arbitrary triangle (this inequality is a strengthened version of Pólya's inequality proposed by Professor Leng Gangsong). | Prove that first rationalize (3.2.1) and make the substitution $a=y+z, b=x+z, c=x+y$, then the original inequality becomes
$$27 \prod(x+y)^{2}\left(\prod(x+y)^{2}-\prod(x-y)^{2}\right)-4096 \sigma_{1}^{3} \sigma_{3}^{3} \geq 0$$
Let the left side of (3.2.2) be denoted as $L_{1}(x, y, z)$. It can be verified that $L_{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,285 |
Example 3: Prove that when $k \geq 1, x>0, y>0, z>0$,
$$\sum \frac{x y z}{y^{3}+z^{3}+k x y z} \leq \frac{3}{2+k}$$ | Prove that after rearranging the above equation, we obtain its equivalent form:
$$\begin{aligned}
L_{3}= & 3 f_{0.4}^{(9)}+3 f_{0.5}^{(9)}+2(k-1) f_{1,1}^{(9)}+(2 k+1) f_{1,2}^{(9)} \\
& +(2 k+1) f_{1.3}^{(9)}+2(5 k+1) f_{1.4}^{(9)}+2\left(k^{2}-k+3\right) f_{2,1}^{(9)}+\left(2 k^{2}+2 k-1\right) f_{2,2}^{(9)}
\end{ali... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,286 |
Example 5 Let $x, y, z$ be positive real numbers. Prove:
$$(x y+y z+z x)\left[\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}}\right] \geqslant \frac{9}{4} .$$
(1996, Iran Mathematical Olympiad) | Prove: The original inequality is equivalent to
$$\begin{array}{l}
f(x, y, z) \\
= 4(x y+y z+z x)\left[(y+z)^{2}(z+x)^{2}+\right. \\
\left.(z+x)^{2}(x+y)^{2}+(x+y)^{2}(y+z)^{2}\right]- \\
\quad 9(x+y)^{2}(y+z)^{2}(z+x)^{2} \\
\geqslant 0 .
\end{array}$$
First, calculate $f(1,0,0), f(1,1,0), f(1,1,1)$, to get $a=0, d=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,290 |
Example 6 Let positive real numbers $x, y, z$ satisfy $xyz \geqslant 1$. Prove:
$$\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}-y^{2}}{y^{5}+z^{2}+x^{2}}+\frac{z^{5}-z^{2}}{z^{5}+x^{2}+y^{2}} \geqslant 0 .$$
(46th IMO) | $$\begin{array}{l}
\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}} \geqslant \frac{x^{5}-x^{2} x y z}{x^{5}+\left(y^{2}+z^{2}\right) x y z} \\
=\frac{x^{4}-x^{2} y z}{x^{4}+\left(y^{2}+z^{2}\right) y z} \\
\text { and } \frac{x^{4}-x^{2} y z}{x^{4}+\left(y^{2}+z^{2}\right) y z}=\frac{2\left(x^{4}-x^{2} y z\right)}{2\left[x^{4}+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,291 |
Example 1 Given that $x, y, z$ are non-negative real numbers, and satisfy $x+y+z=1$. Prove:
$$0 \leqslant x y+y z+z x-2 x y z \leqslant \frac{7}{27} \text {. }$$
(25th IMO) | Analysis: Since the original inequality is not homogeneous, we first homogenize it, i.e., transform it into
$$\begin{array}{l}
0 \leqslant(x y+y z+z x)(x+y+z)-2 x y z \\
\leqslant \frac{7}{27}(x+y+z)^{3} . \\
\text { Let } f(x, y, z) \\
=(x y+y z+z x)(x+y+z)-2 x y z \\
=a g_{3,1}+b g_{3,2}+c g_{3,3} .
\end{array}$$
To... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,295 |
Example 3 Let $a, b, c$ be positive real numbers, and satisfy $abc = 1$. Prove:
$$\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2} \text {. }$$ | Analysis: Directly eliminating the denominator to transform into an integral inequality will lead to higher-degree polynomials, thus encountering greater difficulties. Starting from the known $abc=1$, let's see if we can reduce the degree first.
Make the substitution $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$, the ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,298 |
Example 1 (2001 Austrian-Polish Mathematical Olympiad Problem) Given that $a, b, c$ are the three sides of $\triangle ABC$, prove: $2<$ $\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}-\frac{a^{3}+b^{3}+c^{3}}{a b c} \leqslant 3$. | Prove that the left inequality is equivalent to $(b+c-a)(c+a-b)$ $(a+b-c)>0$, the right inequality is equivalent to $(b+c-a)(c+a$ - b) $(a+b-c) \leqslant a b c$ (Schur's Inequality (Variant 3)) | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,302 |
Example 2 (2009 Greek Mathematical Olympiad) Given that $x, y, z$ are non-negative numbers, and $x+y+z=2$, prove the inequality: $x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+x y z \leqslant 1$ | To prove the homogenization on both sides is equivalent to proving $(x+y+z)$
$$16\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+8 x y z(x+y+z)$$
Since $(x+y+z)^{4}=x^{4}+y^{4}+z^{4}+4\left(x^{3} y+\right.$ $\left.x y^{3}+y^{3} z+y z^{3}+z^{3} x+z x^{3}\right)+6\left(x^{2} y^{2}+y^{2} z^{2}+\right.$ $\left.z^{2} x^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,303 |
Example 3 (2006 Ukrainian Mathematical Olympiad Problem) Given that $a, b, c$ are positive numbers, prove: $3\left(a^{3}+b^{3}+c^{3}+a b c\right) \geqslant$ $4\left(a^{2} b+b^{2} c+c^{2} a\right)$. | $$\begin{array}{l}
\text { Proof: By Schur's Inequality (Variant 1), we have: } \\
a^{3}+b^{3}+c^{3}+3 a b c \geqslant a^{2} b+a b^{2}+b^{2} c+b c^{2}+ \\
c^{2} a+c a^{2}
\end{array}$$
By the AM-GM Inequality, we get $\frac{a^{3}+a^{3}+b^{3}}{3} \geqslant a^{2} b$,
$$\frac{b^{3}+b^{3}+c^{3}}{3} \geqslant b^{2} c, \fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,304 |
Example 4 (2009 Oliforum Mathematical Olympiad Problem) Let $a, b, c$ be positive real numbers, prove: $a+b+c \leqslant \frac{a b}{a+b}+\frac{b c}{b+c} +\frac{c a}{c+a}+\frac{1}{2}\left(\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b}\right)$. | Prove that for $a=x y, b=y z, c=z x$, the original inequality is transformed into
$$\begin{aligned}
& \frac{1}{2}\left(x^{2}+y^{2}+z^{2}\right)+x y z\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right) \\
\geqslant & x y+y z+z x .
\end{aligned}$$
By the Cauchy-Schwarz inequality, we have $\frac{1}{x+y}+\frac{1}{y+z}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,305 |
Example 5 (2003 USA National Training Team Selection Problem) Let $\alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)$, prove the inequality:
$$\begin{array}{l}
\frac{\sin \alpha \sin (\alpha-\beta) \sin (\alpha-\gamma)}{\sin \left(\beta + \gamma\right)}+\frac{\sin \beta \sin (\beta-\alpha) \sin (\beta-\gamma)}{\si... | Prove that since
\[
\sin (x+y) \sin (x-y)=(\sin x \cos y + \cos x \sin y)(\sin x \cos y - \cos x \sin y) = \sin ^{2} x \cos ^{2} y - \sin ^{2} y \cos ^{2} x = \sin ^{2} x (1 - \sin ^{2} y) - \sin ^{2} y (1 - \sin ^{2} x) = \sin ^{2} x - \sin ^{2} y
\]
Therefore, the original inequality is equivalent to
\[
\begin{array... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,306 |
Example 6 (2004 China Western Mathematical Olympiad) Prove that for any positive real numbers $a, b, c$, we have $1<\frac{a}{\sqrt{a^{2}+b^{2}}}+$
$$\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}} \leqslant \frac{3 \sqrt{2}}{2} .$$ | Prove the left inequality first. Let $x=\frac{b^{2}}{c^{2}}, y=\frac{c^{2}}{a^{2}}, z=\frac{a^{2}}{b^{2}}$, then $x, y, z \in \mathbf{R}^{+}, x y z=1$, so we only need to prove $\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}}>1$.
Without loss of generality, assume $x \leqslant y \leqslant z$, let $A=x y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,307 |
Example 7 Let $x, y, z$ be positive real numbers, prove: $\frac{x^{3}+y^{3}+z^{3}}{3 x y z}$ $+\frac{3 \sqrt[3]{x y z}}{x+y+z} \geqslant 2 . \quad(\mathrm{M}$ ircea Lasscu Inequality $)$ | Prove that from the Schur inequality (variant 6) $x+y+z+3 \sqrt[3]{x y z} \geqslant 2(\sqrt{x y}+\sqrt{y z}+\sqrt{z x})$ we get
$$3 \sqrt[3]{x y z} \geqslant 2(\sqrt{x y}+\sqrt{y z}+\sqrt{z x})-(x+y+z)$$
We have
$$\frac{3 \sqrt[3]{x y z}}{x+y+z} \geqslant \frac{2(\sqrt{x y}+\sqrt{y z}+\sqrt{z x})-(x+y+z)}{x+y+z}$$
Th... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,308 |
Example 1.1.1. Let \(a, b, c\) be positive real numbers with sum 3. Prove that
\[
\sqrt{a} + \sqrt{b} + \sqrt{c} \geq ab + bc + ca
\] | Solution. Notice that
$$2(a b+b c+c a)=(a+b+c)^{2}-a^{2}-b^{2}-c^{2}$$
The inequality is then equivalent to
$$\sum_{c y c} a^{2}+2 \sum_{c y c} \sqrt{a} \geq 9$$
which is true by AM-GM because
$$\sum_{c y c} a^{2}+2 \sum_{c y c} \sqrt{a}=\sum_{c y c}\left(a^{2}+\sqrt{a}+\sqrt{a}\right) \geq 3 \sum_{c y c} a=9 .$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,310 |
Example 1.1.10. Let \(a, b, c\) be positive real numbers such that \(abc=1\). Prove that
\[
\sqrt{\frac{a+b}{a+1}}+\sqrt{\frac{b+c}{b+1}}+\sqrt{\frac{c+a}{c+1}} \geq 3
\] | Solution. After applying AM-GM for the three terms on the left hand side expression, we only need to prove that
$$(a+b)(b+c)(c+a) \geq(a+1)(b+1)(c+1)$$
or equivalent by (because $abc=1$)
$$ab(a+b)+bc(b+c)+ca(c+a) \geq \mathbf{a}+b+c+ab+bc+ca$$
According to AM-GM,
$$\begin{array}{l}
2 \mathrm{LHS}+\sum_{c y c} ab=\sum_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,311 |
Example 6.1.3. Let \( x, y, z \) be positive real numbers with sum 3. Find the minimum of the expression \( x^{2} + y^{2} + z^{3} \). | Solution. Let $a$ and $b$ be positive real numbers. By AM-GM inequality, we have
$$\begin{aligned}
x^{2}+a^{2} & \geq 2 a x \\
y^{2}+a^{2} & \geq 2 a y \\
z^{3}+b^{3}+b^{3} & \geq 3 b^{2} z
\end{aligned}$$
Combining these results yields that $x^{2}+y^{2}+z^{3}+2\left(a^{2}+b^{3}\right) \geq 2 a(x+y)+3 b^{2} z$, with e... | 6 a-\left(2 a^{2}+b^{3}\right) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,312 |
Example 6.1.4. Let \(a, b, c\) be three positive constants and \(x, y, z\) three positive variables such that \(a x + b y + c z = x y z\). Prove that if there exists a unique positive number \(d\) such that
\[
\frac{2}{d} = \frac{1}{a + d} + \frac{1}{b + d} + \frac{1}{c + d}
\]
then the minimum of \(x + y + z\) is
\[
\... | Certainly, if the minimum of the above expression is equal to $C$ then the minimum value of $x+y+z$ is equal to $C$, too.
Assume that $m, n, p, m_{1}, n_{1}, p_{1}$ are arbitrary positive real numbers such that $m+n+$ $p=a m_{1}+b n_{1}+c p_{1}=1$. By the weighted AM-GM inequality, we have
$$\begin{array}{c}
x+y+z=m\l... | \sqrt{d(d + a)(d + b)(d + c)} | Algebra | proof | Yes | Yes | inequalities | false | 737,313 |
Example 6.1.5. Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive real numbers. Prove that
$$x_{1}+\sqrt{x_{1} x_{2}}+\ldots+\sqrt[n]{x_{1} x_{2} \ldots x_{n}} \leq e\left(x_{1}+x_{2}+\ldots+x_{n}\right)$$ | Solution. Suppose that $a_{1}, a_{2}, \ldots, a_{n}$ are positive real numbers. According to AM-GM inequality, we have
$$\begin{array}{c}
\sqrt[k]{\left(a_{1} x_{1}\right) \cdot\left(a_{2} x_{2}\right) \cdots\left(a_{k} x_{k}\right)} \leq \frac{a_{1} x_{1}+a_{2} x_{2}+\ldots+a_{k} x_{k}}{k} \\
\Rightarrow \sqrt[k]{x_{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,314 |
Example 6.0.8. Let \(x, y, z\) be positive real numbers and \(xy + yz + zx = 1\). Prove that
\[10x^2 + 10y^2 + z^2 \geq 4\] | Solution. Let's first see a nice, short and a bit magic solution, before we give a general and natural solution. By AM-GM inequality, we deduce that $2 x^{2}+2 y^{2} \geq$ $4 x y, 8 x^{2}+1 / 2 z^{2} \geq 4 x z$ and $8 y^{2}+1 / 2 z^{2} \geq 4 y z$. Adding up these inequalities, we have
$$10 x^{2}+10 y^{2}+z^{2} \geq 4... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,315 |
Example 6.1.1. Let $k$ be a positive real number. Find the minimum of
$$k\left(x^{2}+y^{2}\right)+z^{2},$$
where $x, y, z$ are three positive real numbers such that $x y+y z+z x=1$. | SOLUTION. We separate $k=l+(k-l)$ (with the condition $0 \leq l \leq k)$ and apply AM-GM inequality in the following form
$$\begin{aligned}
l x^{2}+l y^{2} & \geq 2 l x y \\
(k-l) x^{2}+1 / 2 z^{2} & \geq \sqrt{2(k-l)} x z \\
(k-l) y^{2}+1 / 2 z^{2} & \geq \sqrt{2(k-l)} y z
\end{aligned}$$
These results, combined, yie... | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,316 |
Example 6.2.1. Suppose that \(x, y, z\) are three positive real numbers verifying \(x + y + z = 3\). Find the minimum of the expression
\[
x^{4} + 2 y^{4} + 3 z^{4}
\] | Solution. Let $a, b, c$ be three positive real numbers such that $a+b+c=3$. According to Hölder inequality, we obtain
$$\left(x^{4}+2 y^{4}+3 z^{4}\right)\left(a^{4}+2 b^{4}+3 c^{4}\right)^{3} \geq\left(a^{3} x+2 b^{3} y+3 c^{3} z\right)^{4}(\star)$$
We will choose $a, b, c$ such that $a^{3}=2 b^{3}=3 c^{3}=k^{3}$, an... | \frac{\left(3 k^{3}\right)^{4}}{\left(a^{4}+2 b^{4}+3 c^{4}\right)^{3}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,317 |
Example 6.2.2. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers. Prove that
$$\frac{1}{a_{1}}+\frac{2}{a_{1}+a_{2}}+\ldots+\frac{n}{a_{1}+a_{2}+\ldots+a_{n}}<2\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}\right) .$$ | SOLUTION. Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive real numbers (we will determine them at the end). According to the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& \left(a_{1}+a_{2}+\ldots+a_{k}\right)\left(\frac{x_{1}^{2}}{a_{1}}+\frac{x_{2}^{2}}{a_{2}}+\ldots+\frac{x_{k}^{2}}{a_{k}}\right) \geq\left(x_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,318 |
Example 6.2.3. Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive real numbers. Prove that
$$x_{1}^{2}+\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+\ldots+\left(\frac{x_{1}+x_{2}+\ldots+x_{n}}{n}\right)^{2} \leq 4\left(x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}\right)$$ | Solution. Let $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ be positive real numbers. According to the Cauchy-Schwarz inequality, we have
$$\left(\frac{x_{1}^{2}}{\alpha_{1}}+\frac{x_{2}^{2}}{\alpha_{2}}+\ldots+\frac{x_{k}^{2}}{\alpha_{k}}\right)\left(\alpha_{1}+\alpha_{2}+\ldots+\alpha_{k}\right) \geq\left(x_{1}+x_{2}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,319 |
Example 6.2.4. Find the best value of $t=t(n)$ (smallest) for which the following inequality is true for all real numbers $x_{1}, x_{2}, \ldots, x_{n}$
$$x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\ldots+\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{2} \leq t\left(x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}\right)$$
(MYM 2004) | SOLUTION. Let $c_{1}, c_{2}, \ldots, c_{n}$ be positive real numbers (which will be chosen later). According to the Cauchy-Schwarz inequality, we have
$$\left(\sum_{i=1}^{k} x_{i}\right)^{2} \leq S_{k}\left(\sum_{i=1}^{k} \frac{x_{i}^{2}}{c_{i}}\right)$$
where $S_{1}, S_{2}, \ldots, S_{n}$ are determined as
$$S_{k}=\su... | \frac{1}{4 \sin ^{2} \frac{\pi}{2(2 n+1)}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,320 |
Example 7.1.2. Let \(a, b, c\) be positive real numbers. Prove that
\[
\frac{a^{3}}{b^{3}+c^{3}}+\frac{b^{3}}{c^{3}+a^{3}}+\frac{c^{3}}{a^{3}+b^{3}} \geq \frac{a^{2}}{b^{2}+c^{2}}+\frac{b^{2}}{c^{2}+a^{2}}+\frac{c^{2}}{a^{2}+b^{2}}.
\] | Solution. We will solve the general problem for all real numbers $s \geq t \geq 0$
$$\frac{a^{s}}{b^{s}+c^{s}}+\frac{b^{s}}{c^{s}+a^{s}}+\frac{c^{s}}{a^{s}+b^{s}} \geq \frac{a^{t}}{b^{t}+c^{t}}+\frac{b^{t}}{c^{t}+a^{t}}+\frac{c^{t}}{a^{t}+b^{t}} .$$
It's enough to prove that the following function is increasing by der... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,321 |
Example 1.1.11. Let \(a, b, c\) be the side-lengths of a triangle. Prove that
\[
(a+b-c)^{a}(b+c-a)^{b}(c+a-b)^{c} \leq a^{a} b^{b} c^{c} .
\] | Solution. Applying the weighted AM-GM inequality, we conclude that
$$\begin{array}{c}
\sqrt[a+b+c]{\left(\frac{a+b-c}{a}\right)^{a}\left(\frac{b+c-a}{b}\right)^{b}\left(\frac{c+a-b}{c}\right)^{c}} \\
\leq \frac{1}{a+b+c}\left(a \cdot \frac{a+b-c}{a}+b \cdot \frac{b+c-a}{b}+c \cdot \frac{c+a-b}{c}\right)=1 .
\end{array}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,322 |
Example 7.1.4. Prove that $0 \leq a \leq 1 \leq b \leq 3 \leq c \leq 4$ if $a, b, c$ are real numbers satisfying the conditions
$$a \leq b \leq c, a+b+c=6, ab+bc+ca=9$$
(British MO) | Solution. Denote $p=a b c$ and consider the function
$$f(x)=(x-a)(x-b)(x-c)=x^{3}-6 x^{2}+9 x-p$$
We have $f^{\prime}(x)=3 x^{2}-12 x+9=3(x-1)(x-3)$. Therefore $f^{\prime}(x)=0$ or $x=1 \vee x=3$.
Because $f(x)$ has three roots $a \leq b \leq c$, we infer
$$1 \leq b \leq 3, f(1) f(3) \leq 0$$
Note that $f(1)=f(4)=4-p... | proof | Algebra | proof | Yes | Yes | inequalities | false | 737,324 |
Example 7.2.1. Let \(a, b, c\) be positive real numbers. Prove that
\[a^{3}+b^{3}+c^{3}+3 a b c \geq a b(a+b)+b c(b+c)+c a(c+a)\] | SOLUTION. WLOG, assume that $a \geq b \geq c$. Consider the function of $a$: $f(a)=$ $a^{3}+b^{3}+c^{3}+3 a b c-a b(a+b)-b c(b+c)-c a(c+a)$. We have
$$f^{\prime}(a)=3 a^{2}+3 b c-2 a b-b^{2}-2 a c-c^{2}$$
Notice that $f^{\prime \prime}(a)=6 a-2 b-2 c \geq 0$ and $f^{\prime \prime}(b) \geq 0$, so $f^{\prime}(a) \geq f^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,325 |
Example 7.2.2. Let $a, b, c, d$ be positive real numbers such that
$$2(a b+b c+c d+d a+a c+b d)+a b c+b c d+c d a+d a b=16 .$$ | Prove the following inequality
$$a+b+c+d \geq \frac{2}{3}(a b+b c+c d+d a+a c+b d)$$
(Vietnam MO 1996)
SOLUTION. By a similar reasoning as in example 7.1 .3, we deduce that there exist positive real numbers $x, y, z$ for which
$$\sum_{\text {sym }} a=\frac{4}{3} \sum_{\text {sym }} x, \quad \sum_{\text {sym }} a b=2 \s... | proof | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,326 |
Example 7.2.3. Let \(a, b, c\) be non-negative real numbers. Prove that
\[a^{3}+b^{3}+c^{3}+9 a b c+4(a+b+c) \geq 8(a b+b c+c a).\]
(Le Trung Kien) | Solution. We denote
$$f(b)=b^{3}+b(4+9 a c-8 a-8 c)+a^{3}+c^{3}+4(a+c)-8 a c$$
By AM-GM inequality, $\left(a^{3}+4 a\right)+\left(c^{3}+4 c\right) \geq 4 a^{2}+4 c^{2} \geq 8 a c$, so the problem is proved in case $4+9 a c \geq 8(a+c)$, with equality for $a=c=2, b=0$ and $a=b=c=0$. Otherwise, let $x=a+c, y=a c$ then, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,327 |
Example 7.2.4. Suppose $n$ is an integer greater than 2 and that the $n$ positive real numbers $x_{1}, x_{2}, \ldots, x_{n}$ satisfy the condition
$$\left(x_{1}+x_{2}+\ldots+x_{n}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}}\right) \leq(n+\sqrt{10}-3)^{2}$$ | Prove that every 3-uple $\left(x_{i}, x_{j}, x_{k}\right)(1 \leq i < j < k \leq n)$ are real numbers verifying $x_{1}>x_{2}+x_{3}$, then
$$\left(x_{1}+x_{2}+\ldots+x_{n}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}}\right)>(n+\sqrt{10}-3)^{2}$$
Indeed, we will prove it by induction. For $n=3$, the... | proof | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,328 |
Example 7.2.5. Let $a, b, c$ be positive real numbers such that $12 \geq 21 a b + 2 b c + 8 c a$. Prove that
$$\frac{1}{a} + \frac{2}{b} + \frac{3}{c} \leq \frac{15}{2}$$
(Tran Nam Dung, Vietnam TST 2001) | Solution. Although this problem has been solved in the previous part by balancing coefficients, it seems to be more intuitive to give a solution by derivatives. Let \( x = \frac{1}{a}, y = \frac{2}{b}, z = \frac{3}{c} \). We will prove an equivalent problem as follows.
If \( x, y, z > 0 \) and \( 12xyz \geq 2x + 8y + ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,329 |
Let $a, b, c$ be three real numbers. Prove that
$$a^{6}+b^{6}+c^{6}+a^{2} b^{2} c^{2} \geq \frac{2}{3}\left(a^{5}(b+c)+b^{5}(c+a)+c^{5}(a+b)\right)$$ | Solution. According to AM-GM inequality and Schur inequality, we deduce that
$$\begin{aligned}
3 \sum_{c y c} a^{6}+3 a^{2} b^{2} c^{2} & \geq 2 \sum_{c y c} a^{6}+\sum_{c y c} a^{4}\left(b^{2}+c^{2}\right) \\
& =\sum_{c y c}\left(a^{6}+a^{4} b^{2}\right)+\sum_{c y c}\left(a^{6}+a^{4} c^{2}\right) \geq 2 \sum_{c y c} a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,331 |
Example 8.1.1. Let \(a, b, c\) be non-negative real numbers with sum 2. Prove that
$$a^{4}+b^{4}+c^{4}+a b c \geq a^{3}+b^{3}+c^{3}$$ | SoluTION. According to the fourth degree-Schur inequality, we have
$$a^{4}+b^{4}+c^{4}+a b c(a+b+c) \geq a^{3}(b+c)+b^{3}(c+a)+c^{3}(a+b)$$
or equivalently
$$2\left(a^{4}+b^{4}+c^{4}\right)+a b c(a+b+c) \geq\left(a^{3}+b^{3}+c^{3}\right)(a+b+c)$$
Putting $a+b+c=2$ into the last inequality, we get the desired result. E... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,332 |
Example 1.1.12. Let \(a, b, c\) be non-negative real numbers with sum 2. Prove that
\[a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} \leq 2\] | Solution. We certainly have
$$(a b+b c+c a)\left(a^{2}+b^{2}+c^{2}\right) \geq \sum_{c y c} a^{3}(b+c)=\sum_{c y c} a b\left(a^{2}+b^{2}\right) \geq 2 \sum_{c y c} a^{2} b^{2}$$
Applying AM-GM, and using $a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=4$, we deduce that
$$2(a b+b c+c a)\left(a^{2}+b^{2}+c^{2}\right) \leq 4 \Rightar... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,333 |
Example 8.1.2. Let $a, b, c$ be positive real numbers. Prove that
$$\frac{a^{2}}{\sqrt{(b+c)\left(b^{3}+c^{3}\right)}}+\frac{b^{2}}{\sqrt{(c+a)\left(c^{3}+a^{3}\right)}}+\frac{c^{2}}{\sqrt{(a+b)\left(a^{3}+b^{3}\right)}} \geq \frac{8}{2}$$
(Pham Kim Hung) | Solution. By Hölder inequality, we deduce that
$$\left(\sum_{c y c} \frac{a^{2}}{\sqrt{(b+c)\left(b^{3}+c^{3}\right)}}\right)^{2}\left(\sum_{c y c} a^{2}(b+c)\left(b^{3}+c^{3}\right)\right) \geq\left(\sum_{c y c} a^{2}\right)^{3} .$$
So it is enough to prove that
$$\begin{array}{c}
4\left(\sum_{c y c} a^{2}\right)^{3}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,334 |
Example 8.1.3. Suppose that \(a, b, c\) are non-negative real numbers. Prove that
\[
\frac{a^{2}}{2 b^{2}-b c+2 c^{2}}+\frac{b^{2}}{2 c^{2}-c a+2 a^{2}}+\frac{c^{2}}{2 a^{2}-a b+2 b^{2}} \geq 1
\]
(Vasile Cirtoaje) | Solution. According to Cauchy-Schwarz inequality, we deduce that
$$\sum_{c y c} \frac{a^{2}}{2 b^{2}-b c+2 c^{2}} \geq \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a^{2}\left(2 b^{2}-b c+2 c^{2}\right)+b^{2}\left(2 c^{2}-c a+2 a^{2}\right)+c^{2}\left(2 a^{2}-a c+2 b^{2}\right)}$$
It suffices to prove that
$$\left(\sum_{c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,335 |
Example 8.1.4. Let \(a, b, c\) be non-negative real numbers. Prove that
\[
\frac{a^{3}}{b^{2}-b c+c^{2}}+\frac{b^{3}}{c^{2}-c a+a^{2}}+\frac{c^{3}}{a^{2}-a b+b^{2}} \geq a+b+c.
\] | Solution. Applying Cauchy-Schwarz inequality, we have
$$\sum_{c y c} \frac{a^{3}}{b^{2}-b c+c^{2}}=\sum_{c y c} \frac{a^{4}}{a\left(b^{2}-b c+c^{2}\right)} \geq \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{\sum_{c y c} a\left(b^{2}-b c+c^{2}\right)}$$
It remains to prove that
$$\left(\sum_{c y c} a^{2}\right)^{2} \geq\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,336 |
Example 8.1.5. Let $a, b, c$ be non-negative real numbers. Prove that
$$a^{2} \sqrt{b^{2}-b c+c^{2}}+b^{2} \sqrt{c^{2}-c a+a^{2}}+c^{2} \sqrt{a^{2}-a b+b^{2}} \leq a^{3}+b^{3}+c^{3}$$ | Solution. Applying AM-GM inequality, we have
$$\sum_{c y c} a^{2} \sqrt{b^{2}-b c+c^{2}}=a \sqrt{a^{2}\left(b^{2}-b c+c^{2}\right)} \leq \frac{1}{2} \sum_{c y c} a\left(a^{2}+b^{2}+c^{2}-b c\right)$$
Then, by the third degree-Schur inequality we get that
$$2 \sum_{c y c} a^{3}-\sum_{c y c} a\left(a^{2}+b^{2}+c^{2}-b c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,337 |
Example 8.1.6. Let \(a, b, c\) be non-negative real numbers. Prove that
\[
\frac{a^{3}}{\sqrt{b^{2}-b c+c^{2}}}+\frac{b^{3}}{\sqrt{c^{2}-c a+a^{2}}}+\frac{c^{3}}{\sqrt{a^{2}-a b+b^{2}}} \geq a^{2}+b^{2}+c^{2} .
\] | Solution. Applying Cauchy-Schwarz inequality, we deduce that
$$\sum_{\text {cyc }} \frac{a^{3}}{\sqrt{b^{2}-b c+c^{2}}} \geq \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a \sqrt{b^{2}-b c+c^{2}}+b \sqrt{c^{2}-c a+a^{2}}+c \sqrt{a^{2}-a b+b^{2}}}$$
So it is enough to prove that
$$\sum_{c y c} a \sqrt{b^{2}-b c+c^{2}} \leq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,338 |
Example 8.2.1. Let \(a, b, c\) be positive real numbers with sum 2. Prove that
\[a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} + a b c \leq 1\]
(Pham Kim Hung) | Solution. Because $a+b+c=2$, the inequality is equivalent to
$$(a b+b c+c a)^{2} \leq 1+3 a b c$$
Denote $x=a b+b c+c a$ and $y=a b c$. We are done if $x \leq 1$. Otherwise, $x \geq 1$, and by AM-GM inequality, we deduce that
$$\prod_{c y c}(a+b-c) \leq \prod a b c \Rightarrow 8 \prod_{c y c}(1-a) \leq \prod_{c y c} a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,339 |
Example 8.2.2. Suppose that \(a, b, c\) are non-negative real numbers such that \(a^{2} + b^{2} + c^{2} = 1\). Prove the inequality
\[
a + b + c \leq \sqrt{2} + \frac{9abc}{4}
\] | Solution. We denote $x=a+b+c, y=ab+bc+ca$ and $z=abc$. By the fourth degree-Schur inequality, we have
$$\begin{array}{c}
\sum_{cyc} a^{4} + abc \sum_{cyc} a \geq \sum_{cyc} a^{3}(b+c) \\
\Leftrightarrow \left(\sum_{cyc} a^{2}\right)^{2} - 2\left(\sum_{cyc} ab\right)^{2} + 6abc\left(\sum_{cyc} a\right) \geq \left(\sum_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,340 |
Example 8.2.3. Let $a, b, c$ be positive real numbers satisfying $a b c=1$. Prove that
$$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a} \leq \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$$
(Bulgarian MO 1998) | SOLUTION. Denote \( S = \sum_{cyc} a \), \( P = \sum_{cyc} ab \), \( Q = abc \). By some calculations, we get that
\[
\begin{array}{c}
\mathrm{LHS} = \sum_{cyc} \frac{1}{S+1-a} = \frac{S^2 + 4S + 3 + P}{S^2 + 2S + PS + P} \\
\mathrm{RHS} = \sum_{cyc} \frac{1}{2+a} = \frac{12 + 4S + P}{9 + 4S + 2P}
\end{array}
\]
So it... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,341 |
Example 8.2.4. Let \(a, b, c\) be positive real numbers. Prove that
\[
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{a b c}{2\left(a^{3}+b^{3}+c^{3}\right)} \geq \frac{5}{3} .
\] | SOLUTION. WLOG, we may assume that $a+b+c=3$. Denote $\mathrm{x}=a b+b c+c a$ and $y=a b c$. Then we have
$$\begin{array}{c}
\frac{a b c}{a^{3}+b^{3}+c^{3}}=\frac{y}{27+3 y-9 x} \\
\sum_{c y c} \frac{a}{b+c}=\frac{27+3 y-6 x}{3 x-y}
\end{array}$$
We need to prove that
$$\frac{27+3 y-6 x}{3 x-y}+\frac{y}{2(27+3 y-9 x)}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,342 |
Example 8.2.5. Let $a, b, c$ be real numbers with sum 3. Prove that
$$\left(1+a+a^{2}\right)\left(1+b+b^{2}\right)\left(1+c+c^{2}\right) \geq 9(a b+b c+c a)$$
(Pham Kim Hung) | Solution. We denote
$$x=a+b+c, y=ab+bc+ca, z=abc .$$
According to the hypothesis, $x=3$, so we can rewrite the inequality to
$$\begin{array}{c}
z^{2}+z+1+\sum_{\text{sym}}\left(a+a^{2}\right)+\sum_{\text{sym}} ab+\sum_{\text{sym}} a^{2}b^{2}+abc\left(\sum_{\text{sym}} a+\sum_{\text{sym}} ab\right)+\sum_{\text{sym}} a^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,343 |
Example 1.1.13. Let \(a, b, c, d\) be positive real numbers. Prove that
\[
\frac{1}{a^{2}+a b}+\frac{1}{b^{2}+b c}+\frac{1}{c^{2}+c d}+\frac{1}{d^{2}+d a} \geq \frac{4}{a c+b d}
\] | Solution. Notice that
$$\frac{a c+b d}{a^{2}+a b}=\frac{a^{2}+a b+a c+b d}{a(a+b)}-1=\frac{a(a+c)+b(d+a)}{a(a+b)}-1=\frac{a+c}{a+b}+\frac{b(d+a)}{a(a+d)}-1 .$$
According to AM-GM inequality, we get that
$$(a c+b d)\left(\sum_{c y c} \frac{1}{a^{2}+a b}\right)=\left(\sum_{c y c} \frac{a+c}{a+b}\right)+\left(\sum_{c y c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,344 |
Example 8.2.6. Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that
$$\frac{1}{1+3 a}+\frac{1}{1+3 b}+\frac{1}{1+3 c}+\frac{1}{1+a+b+c} \geq 1$$
(Pham Kim Hung) | Solution. We denote \(x=a+b+c\) and \(y=ab+bc+ca\). Then the inequality can be rewritten to
\[
\begin{array}{c}
\frac{3+6x+9y}{28+3x+9y} + \frac{1}{1+x} \geq 1 \Leftrightarrow \frac{1}{1+x} \geq \frac{25-3x}{28+3x+9y} \\
\Leftrightarrow 3x^2 - 19x + 9y + 3 \geq 0
\end{array}
\]
Denote \(z = \sqrt{\frac{x}{3}}\). Becau... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,345 |
Example 8.2.7. Let \(a, b, c\) be non-negative real numbers with sum 1. Prove that
\[
\frac{a b + b c + c a}{a^{2} + b^{2} + c^{2} + 16 a b c} \geq 8\left(a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}\right)
\]
(Pham Kim Hung, MYM) | Solution. Denote \(x=4(ab+bc+ca)\) and \(y=8abc\) then we obtain
\[a^{2}+b^{2}+c^{2}=1-\frac{x}{2} ; \quad a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}=\frac{x^{2}}{16}-\frac{y}{4}\]
We can rewrite the inequality to the form
\[\begin{aligned}
& 2 x \geq(4-2 x+8 y)\left(x^{2}-4 y\right) \\
\Leftrightarrow & x(x-1)^{2} \geq 4 y((x-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,346 |
Example 8.2.8. Let \(a, b, c\) be non-negative real numbers with sum 2. Prove that
\[a^{2}+b^{2}+c^{2} \geq 2\left(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}+4 a^{2} b^{2} c^{2}\right)\]
(Pham Kim Hung) | SOLUTION. Write $p=a b+b c+c a$ and $q=a b c$. The inequality can be rewritten to
$$\begin{aligned}
(a+b+c)^{2}-2(a b+b c+c a) \geq & 2(a b+b c+c a)^{3}-6 a b c(a+b+c)(a b+b c+c a)+14 a^{2} b^{2} c^{2} \\
& \Leftrightarrow 2-p \geq p^{3}-6 p q+7 q^{2}
\end{aligned}$$
Let $r=\max \left\{0, \frac{8 p-8}{9}\right\}$, the... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,347 |
Example 8.3.1. Suppose that \(a, b, c\) are three real numbers satisfying \(a^{2}+b^{2}+c^{2}=3\).
Prove the following inequality
\[
a^{3}(b+c)+b^{3}(c+a)+c^{3}(a+b) \leq 6
\] | Solution. Certainly, the inequality is non-homogeneous. However, the condition $a^{2}+$ $b^{2}+c^{2}=3$ can help change it to homogeneous as
$$a^{3}(b+c)+b^{3}(c+a)+c^{3}(a+b) \leq \frac{2}{3}\left(a^{2}+b^{2}+c^{2}\right)^{2}$$
Rewrite this inequality to
$$2 \sum_{c y c} a^{4}+4 \sum_{c y c} a^{2} b^{2} \geq 3 \sum_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,348 |
Example 8.3.2. Let \(a, b, c\) be non-negative real numbers. Prove that
\[
\sqrt{\frac{a b + b c + c a}{3}} \leq \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}}
\] | Solution. WLOG, suppose that $a b+b c+c a=3$. By AM-GM inequality, we deduce that $a+b+c \geq 3$ and $a b c \leq 1$. Therefore
$$\begin{array}{c}
(a+b)(b+c)(c+a)=(a+b+c)(a b+b c+c a)-a b c=3(a+b+c)-a b c \geq 8 \\
\Rightarrow \sqrt{\frac{a b+b c+c a}{3}} \leq 1 \leq \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}}
\end{array}$$
Th... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,349 |
Example 8.3.3. Let $a, b, c$ be non-negative real numbers. Prove that
$$\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(2 b+c+a)^{2}}{2 b^{2}+(c+a)^{2}}+\frac{(2 c+a+b)^{2}}{2 c^{2}+(a+b)^{2}} \leq 8$$
(USA MO 2003) | Solution. By normalizing the expression with $a+b+c=3$, we reduce the left expression to a simpler form
$$\frac{(3+a)^{2}}{2 a^{2}+(3-a)^{2}}+\frac{(3+b)^{2}}{2 b^{2}+(3-b)^{2}}+\frac{(3+c)^{2}}{2 c^{2}+(3-c)^{2}}$$
Notice that
$$\begin{aligned}
\frac{3(3+a)^{2}}{2 a^{2}+(3-a)^{2}} & =\frac{a^{2}+6 a+9}{a^{2}-2 a+3}=1... | 8 | Inequalities | proof | Yes | Yes | inequalities | false | 737,350 |
Example 8.3.4. Let $a, b, c$ be non-negative real numbers. Prove that
$$\frac{(b+c-a)^{2}}{(b+c)^{2}+a^{2}}+\frac{(c+a-b)^{2}}{(c+a)^{2}+b^{2}}+\frac{(a+b-c)^{2}}{(a+b)^{2}+c^{2}} \geq \frac{3}{5}$$ | Solution. WLOG, we may suppose that $a+b+c=3$. The inequality becomes
$$\sum_{cyc} \frac{(3-2a)^{2}}{a^{2}+(3-a)^{2}} \geq \frac{3}{5} \Leftrightarrow \sum_{cyc} \frac{1}{2a^{2}-6a+9} \leq \frac{3}{5}$$
Just notice that
$$\begin{array}{c}
\sum_{cyc}\left(\frac{5}{2a^{2}-6a+9}-1\right)=\sum_{cyc} \frac{2(a-1)(a-2)}{2a^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,351 |
Example 8.3.5. Let \(a, b, c\) be positive real numbers. Prove that
\[
\frac{(2 a+b+c)^{2}}{4 a^{3}+(b+c)^{3}}+\frac{(2 b+c+a)^{2}}{4 b^{3}+(c+a)^{3}}+\frac{(2 c+a+b)^{2}}{4 c^{3}+(a+b)^{3}} \leq \frac{12}{a+b+c} .
\]
(Pham Kim Hung) | Solution. Suppose that $a+b+c=3$. The problem becomes
$$\sum_{c y c} \frac{(3+a)^{2}}{4 a^{3}+(3-a)^{3}} \leq 4$$
Notice that
$$\begin{aligned}
& \frac{(3+a)^{2}}{4 a^{3}+(3-a)^{3}}-\frac{4}{3}=\frac{(a-1)\left(-4 a^{2}-15 a+27\right)}{4 a^{3}+(3-a)^{3}} \\
= & (a-1)\left(\frac{2}{3}+\frac{(a-1)\left(-2 a^{2}-12 a-9\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,352 |
Example 8.3.7. Let $a, b, c$ be non-negative real numbers. Prove that
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{3 \sqrt[3]{a b c}}{2(a+b+c)} \geq 2$$
(Pham Kim Hung) | Solution. Applying Cauchy-Schwarz inequality, we get
$$\sum_{c y c} \frac{a}{b+c} \geq \frac{(a+b+c)^{2}}{2(a b+b c+c a)}$$
We normalize $a+b+c=1$ and prove that
$$\frac{1}{2 x}+\frac{3}{2} \sqrt[3]{a b c} \geq 2$$
where $x=a b+b c+c a \leq \frac{1}{3}$. If $x \leq \frac{1}{4}$, we are done immediately. Otherwise, by ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,353 |
Example 8.3.8. Let $a, b, c$ be non-negative real numbers. Prove that
$$\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}} \geq \frac{9}{4(a b+b c+c a)}$$
(Iran TST 1996) | SOLUTION. We normalize $a b+b c+c a=1$. The inequality becomes
$$4 \sum_{c y c}(a+b)^{2}(a+c)^{2} \geq 9(a+b)^{2}(b+c)^{2}(c+a)^{2}$$
or
$$4\left(1+a^{2}\right)^{2}+4\left(1+b^{2}\right)^{2}+4\left(1+c^{2}\right)^{2} \geq 9(a+b+c-a b c)^{2}$$
Denote $s=a+b+c$. We can rewrite the inequality in terms of $s$ and $a b c$ ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,354 |
Example 1.1.14. Let \(a, b, c, d, e\) be non-negative real numbers such that \(a + b + c + d + e = 5\). Prove that
\[
abc + bcd + cde + dea + eab \leq 5
\] | Solution. Without loss of generality, we may assume that \( e = \min(a, b, c, d, e) \). According to AM-GM, we have
\[
\begin{aligned}
a b c + b c d + c d e + d e a + e a b & = e(a+c)(b+d) + b c(a+d-e) \\
& \leq e\left(\frac{a+c+b+d}{2}\right)^{2} + \left(\frac{b+c+a+d-e}{3}\right)^{2} \\
& = \frac{e(5-e)^{2}}{4} + \fr... | (e-1)^{2}(e+8) \geq 0 | Inequalities | proof | Yes | Yes | inequalities | false | 737,355 |
Example 8.3.9. Let \(a, b, c, d\) be positive real numbers. Prove that
\[
\begin{array}{c}
\frac{a b c}{(d+a)(d+b)(d+c)}+\frac{b c d}{(a+b)(a+c)(a+d)}+ \\
+\frac{c d a}{(b+a)(b+c)(b+d)}+\frac{d a b}{(c+a)(c+b)(c+d)} \geq \frac{1}{2}
\end{array}
\]
(Nguyen Van Thach) | SOLUTION. Denote \( x = \frac{1}{a}, y = \frac{1}{b}, z = \frac{1}{c}, t = \frac{1}{d} \), the inequality becomes
$$\begin{array}{l}
\frac{x^{3}}{(x+y)(x+z)(x+t)}+\frac{y^{3}}{(y+x)(y+z)(y+t)}+ \\
+\frac{z^{3}}{(z+x)(z+y)(z+t)}+\frac{t^{3}}{(t+x)(t+y)(t+z)} \geq \frac{1}{2}
\end{array}$$
WLOG, assume that \( x + y + z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,356 |
Example 8.4.1. Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that
$$\frac{1}{3 a^{2}+(a-1)^{2}}+\frac{1}{3 b^{2}+(b-1)^{2}}+\frac{1}{3 c^{2}+(c-1)^{2}} \geq 1$$
(Le Huu Dien Khue) | Solution. We want to find a real constant $k$ such that
$$\frac{1}{3 a^{2}+(a-1)^{2}} \geq \frac{1}{3}+k \ln a$$
If there exists such a valid number $k$, we can conclude (notice $\ln a+\ln b+\ln c=0$ )
$$\sum_{c y c} \frac{1}{3 a^{2}+(a-1)^{2}} \geq 1+k\left(\sum_{c y c} \ln a\right)=1$$
Denote
$$f(x)=\frac{1}{3 x^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,357 |
Example 8.4.2. Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that
$$\frac{1}{a^{2}-a+1}+\frac{1}{b^{2}-b+1}+\frac{1}{c^{2}-c+1} \leq 3$$
(Vu Dinh Quy) | Solution. First, we will prove that
$$f(x)=\frac{1}{x^{2}-x+1}+\ln x-1 \geq 0 \forall x \in(0,1.8] .$$
Indeed, we have
$$f^{\prime}(x)=\frac{-2 x+1}{\left(x^{2}-x+1\right)^{2}}+\frac{1}{x}=\frac{(x-1)\left(x^{3}-x^{2}-1\right)}{\left(x^{2}-x-1\right)^{2}}$$
The equation $x^{3}=x^{2}+1$ has exactly one real root in $(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,358 |
Example 8.4.4. Suppose that $a, b, c, d$ are positive real numbers satisfying $a b c d=1$.
Prove that
$$\frac{1+a}{1+a^{2}}+\frac{1+b}{1+b^{2}}+\frac{1+c}{1+c^{2}}+\frac{1+d}{1+d^{2}} \leq 4$$
(Vasile Cirtoaje) | Solution. Consider the following function with $x>0$
$$f(x)=\frac{1+x}{1+x^{2}}+\frac{\ln x}{2}-1$$
We obtain
$$f^{\prime}(x)=\frac{(x-1)\left(x^{3}-x^{2}-3 x-1\right)}{2 x\left(1+x^{2}\right)^{2}}$$
Since the equation $x^{3}=x^{2}+3 x+1$ has exactly one real root $x_{0}$ and $4 \geq x_{0}>1$ it's easy to get
$$\max ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,360 |
Example 8.4.5. Suppose that $a, b, c, d, e, f$ are six positive real numbers satisfying $a b c d e f=1$. Prove the following inequality
$$\frac{a-1}{a^{2}+a+1}+\frac{b-1}{b^{2}+b+1}+\frac{c-1}{c^{2}+c+1}+\frac{d-1}{d^{2}+d+1}+\frac{e-1}{e^{2}+e+1}+\frac{f-1}{f^{2}+f+1} \leq 0$$
(Pham Kim Hung) | Solution. Consider the following function with $x>0$
$$f(x)=\frac{x-1}{x^{2}+x+1}-\frac{\ln x}{3} .$$
Notice that
$$f^{\prime}(x)=\frac{(x-1)\left(-x^{3}-6 x^{2}-3 x+1\right)}{3 x\left(x^{2}+x+1\right)^{2}} .$$
The equation $x^{3}+6 x^{2}+3 x=1$ has exactly one real root $x_{0}$ and clearly, $x_{0}>\frac{1}{5}$. Ther... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,361 |
Example 8.4.6. Suppose that \(a, b, c, d, e, f, g\) are positive real numbers such that \(a + b + c + d + e + f + g = 7\). Prove that
\[
\left(a^{2} - a + 1\right)\left(b^{2} - b + 1\right)\left(c^{2} - c + 1\right)\left(d^{2} - d + 1\right)\left(e^{2} - e + 1\right)\left(f^{2} - f + 1\right)\left(g^{2} - g + 1\right) ... | Solution. The inequality is equivalent to
$$\sum_{c y c} \ln \left(a^{2}-a+1\right) \geq 0$$
Consider the function $f(x)=\ln \left(x^{2}-x+1\right)-x+1$. We have
$$f^{\prime}(x)=\frac{(x-1)(2-x)}{x^{2}+x+1}$$
It's easy to get that
$$\min _{0 \leq x \leq 2.75} f(x)=\min \{f(1), f(2.75)\}=0$$
If all $a, b, c, d, e, f,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,362 |
Example 8.4.7. Let \(a, b, c, d\) be positive real numbers such that \(a+b+c+d=4\).
Prove that
\[
\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}} \geq a^{2}+b^{2}+c^{2}+d^{2}
\] | Solution. Consider the following function of positive variable $x$
$$f(x)=\frac{1}{x^{2}}-x^{2}+4 x-4$$
8.0. A note on symmetric inequalities
141
We clearly have
$$f^{\prime}(x)=\frac{-2}{x^{3}}-2 x+4=\frac{-2(x-1)\left(x^{3}-x^{2}-x-1\right)}{x^{3}}$$
The equation $f^{\prime}(x)=0$ has exactly two positive real root... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,363 |
Problem 2. Suppose that $a, b, c$ are positive real numbers. Prove that
$$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right) \geq \frac{9}{1+a b c}$$ | Solution. By Rearrangement inequality,
$$\frac{1}{a(1+a)}+\frac{1}{b(1+b)}+\frac{1}{c(1+c)} \geq \frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{1}{a(1+b)}$$
Hence the inequality will be proved if these relations are fulfilled
$$\sum_{c y c} \frac{1}{b(1+c)} \geq \frac{3}{1+a b c} ; \sum_{c y c} \frac{1}{a(1+c)} \geq \frac{3}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,365 |
Example 1.1.15. Let \(a, b, c, d\) be positive real numbers. Prove that
\[
\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)^{2} \geq \frac{1}{a^{2}}+\frac{4}{a^{2}+b^{2}}+\frac{9}{a^{2}+b^{2}+c^{2}}+\frac{16}{a^{2}+b^{2}+c^{2}+d^{2}}
\] | Solution. We have to prove that
$$\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}+\sum_{sym} \frac{2}{a b} \geq \frac{4}{a^{2}+b^{2}}+\frac{9}{a^{2}+b^{2}+c^{2}}+\frac{16}{a^{2}+b^{2}+c^{2}+d^{2}}$$
1.0. AM-GM inequality
25
By AM-GM, we have
$$\begin{array}{l}
\frac{2}{a b} \geq \frac{4}{a^{2}+b^{2}} \\
\frac{2}{a c}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,366 |
Problem 3. Let $a, b, c$ be positive real numbers satisfying $a^{2}+b^{2}+c^{2}=3$. Prove that
$$\frac{a}{a^{2}+2 b+3}+\frac{b}{b^{2}+2 c+3}+\frac{c}{c^{2}+2 a+3} \leq \frac{1}{2}$$ | SOLUTION. We certainly have $a^{2}+1 \geq 2 a, b^{2}+1 \geq 2 b, c^{2}+1 \geq 2 c$, therefore
$$\sum_{i c y c} \frac{a}{a^{2}+2 b+3} \geq \sum_{i c y c} \frac{a}{2(a+b+1)}$$
It remains to prove that
$$\sum_{c y c} \frac{a}{a+b+1} \leq 1 \Leftrightarrow \sum_{c y c} \frac{b+1}{a+b+1} \geq 2$$
Notice that $a^{2}+b^{2}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,367 |
Problem 4. Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive real numbers and $x_{1} x_{2} \ldots x_{n}=1$. Prove that
$$x_{1}+x_{2}+\ldots+x_{n} \geq \frac{2}{1+x_{1}}+\frac{2}{1+x_{2}}+\ldots+\frac{2}{1+x_{n}}$$ | Solution. Because $2-\frac{2}{1+x_{i}}=\frac{2 x_{i}}{1+x_{i}}$, the problem can be rewritten to
$$\sum_{c y c} x_{i}+\sum_{c y c} \frac{2 x_{i}}{x_{i}+1} \geq 2 n$$
According to AM-GM inequality, we conclude that
$$\begin{array}{c}
-2 n+\sum_{c y c} x_{i}+\sum_{c y c} \frac{2 x_{i}}{x_{i}+1}=-2 n+\sum_{c y c}\left(\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,368 |
Problem 5. Prove that for all positive real numbers $a, b, c \in[1,2]$, we have
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 6\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right) .$$ | 9.0. Problems and Solutions
SOLUTION. Instead of the condition $a, b, c \in [1,2]$, we will prove this inequality with a stronger condition that $a, b, c$ are the side lengths of a triangle. Using the identity
$$\left(\sum_{c y c} a\right)\left(\sum_{c y c} \frac{1}{a}\right)-9=\sum_{c y c} \frac{(a-b)^{2}}{a b}, \qua... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,369 |
Problem 6. Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Prove that
$$a^{2}+b^{2}+c^{2} \geq \frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}$$ | Solution. (Cauchy reverse) By AM-GM, we deduce that
$$\sum_{c y c} \frac{2+a}{2+b}=\sum_{c y c} \frac{2+a}{2}-\sum_{c y c} \frac{b(2+a)}{2(2+b)} \geq \frac{9}{2}-\frac{3 \sqrt[3]{a b c}}{2}$$
So it's enough to prove that
$$\begin{aligned}
& \sum_{c y c} a^{2}+\frac{3 \sqrt[3]{a b c}}{2} \geq \frac{(a+b+c)^{2}}{2} \\
\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,370 |
Problem 7. Let \(x, y, z\) be non-negative real numbers with sum 3. Prove that
\[
\sqrt{\frac{x}{1+2 y z}}+\sqrt{\frac{y}{1+2 z x}}+\sqrt{\frac{z}{1+2 x y}} \geq \sqrt{3}
\] | Solution. According to Cauchy-Schwarz inequality, we deduce that
$$\begin{array}{c}
\sum_{c y c} \sqrt{\frac{x}{1+2 y z}}=\sum_{c y c} \frac{x^{2}}{\sqrt{x} \sqrt{x^{2}+2 x^{2} y z}} \\
\geq \frac{(x+y+z)^{2}}{\sqrt{x} \cdot \sqrt{x^{2}+2 x^{2} y z}+\sqrt{y} \cdot \sqrt{y^{2}+2 y^{2} z x}+\sqrt{z} \cdot \sqrt{z^{2}+2 z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,371 |
Problem 8. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers and $a_{1} a_{2} \ldots a_{n}=1$. Prove that
$$\sqrt{1+a_{1}^{2}}+\sqrt{1+a_{2}^{2}}+\ldots+\sqrt{1+a_{n}^{2}} \leq \sqrt{2}\left(a_{1}+a_{2}+\ldots+a_{n}\right)$$ | SOLUTION. From the obvious inequality $(\sqrt{x}-1)^{4} \geq 0$, we see that
$$\frac{1+x^{2}}{2} \leq(x-\sqrt{x}+1)^{2} \Rightarrow \sqrt{\frac{1+x^{2}}{2}}+\sqrt{x} \leq 1+x$$
According to this result, we have of course
$$\sum_{i=1}^{n} \sqrt{1+a_{i}^{2}} \leq \sqrt{2} \sum_{i=1}^{n} a_{i}+\sqrt{2}\left(n-\sum_{i=1}^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,372 |
Problem 9. Let $a, b, c, k$ be positive real numbers. Prove that
$$\frac{a+k b}{a+k c}+\frac{b+k c}{b+k a}+\frac{c+k a}{c+k b} \leq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$$ | Solution. We denote
$$X=\frac{1+k \cdot \frac{a}{b}}{1+k}, Y=\frac{1+k \cdot \frac{b}{c}}{1+k}, Z=\frac{1+k \cdot \frac{c}{a}}{1+k}$$
According to Hölder's inequality, we get
$$\prod_{c y c}\left(1+\frac{k a}{b}\right) \geq(1+k)^{3}$$
or equivalently $X Y Z \geq 1$. Now rewrite the inequality into the following form
$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,373 |
Problem 10. Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that
$$\frac{1}{(a+1)(a+2)}+\frac{1}{(b+1)(b+2)}+\frac{1}{(c+1)(c+2)} \geq \frac{1}{2}$$ | Solution. By hypothesis \(abc=1\), so there are three positive real numbers \(x, y, z\) for which \(a=\frac{yz}{x^2}, b=\frac{zx}{y^2}, c=\frac{xy}{z^2}\). The inequality becomes
\[
\sum_{cyc} \frac{x^4}{(x^2+yz)(2x^2+yz)} \geq \frac{1}{2}
\]
According to the Cauchy-Schwarz inequality, we deduce that
\[
\text{LHS} \ge... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,374 |
Problem 11. Let $a, b, c, d$ be non-negative real numbers such that $a+b+c+d=4$.
Prove that
$$a^{2}+b^{2}+c^{2}+d^{2}-4 \geq 4(a-1)(b-1)(c-1)(d-1)$$ | Solu'TION. Applying AM-GM inequality, we get
$$\begin{aligned}
a^{2}+b^{2}+c^{2}+d^{2}-4 & =(a-1)^{2}+(b-1)^{2}+(c-1)^{2}+(d-1)^{2} \\
& \geq 4 \sqrt{|(a-1)(b-1)(c-1)(d-1)|}
\end{aligned}$$
It suffices to consider the inequality in the case \(a \geq b \geq 1 \geq c \geq d\) (in order to have \((a-1)(b-1)(c-1)(d-1) \ge... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,375 |
Problem 12. Let \( x, y, z \) be distinct real numbers. Prove that
\[
\frac{x^{2}}{(x-y)^{2}}+\frac{y^{2}}{(y-z)^{2}}+\frac{z^{2}}{(z-x)^{2}} \geq 1
\] | Solution. We have
$$\begin{array}{c}
\sum_{c y c}\left(1-\frac{x}{z}\right)^{2}\left(1-\frac{z}{y}\right)^{2}-\left(1-\frac{y}{x}\right)^{2}\left(1-\frac{z}{y}\right)^{2}\left(1-\frac{x}{z}\right)^{2}= \\
=\sum_{c y c}\left(1-\frac{x}{z}-\frac{z}{y}+\frac{x}{y}\right)^{2}-\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,376 |
Example 1.1.16. Determine the least $M$ for which the inequality
$$\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \leq M\left(a^{2}+b^{2}+c^{2}\right)^{2}$$
holds for all real numbers $a, b$ and $c$. | SOLUTION. Denote \( x = a - b, y = b - c, z = c - a \) and \( s = a + b + c \). Rewrite the inequality in the following form
\[ 9|s x y z| \leq M\left(s^{2} + x^{2} + y^{2} + z^{2}\right)^{2} \]
in which \( s, x, y, z \) are arbitrary real numbers with \( x + y + z = 0 \).
The fact that \( s \) is an independent variab... | \frac{9 \sqrt{2}}{32} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,377 |
Problem 13. Let $a, b, c, d$ be non-negative real numbers with sum 3. Prove that
$$a b(b+c)+b c(c+d)+c d(d+a)+d a(a+b) \leq 4$$ | Solution. WLOG, we may assume that $b+d \leq a+c$. We have
$$\begin{array}{l}
a b(b+c)+b c(c+d)+c d(d+a)+d a(a+b)= \\
=(a+c)(b c+d a)+(b+d)(a b+c d)= \\
=(a+c)((a+c)(b+d)-(a b+c d))+(b+d)(a b+c d)= \\
=(a+c)^{2}(b+d)+(a b+c d)(b+d-a-c) \leq(a+c)^{2}(b+d)
\end{array}$$
Finally, AM-GM inequality shows that
$$2(a+c)^{2}(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,378 |
Problem 14. Let $a_{1}, a_{2}, \ldots, a_{n}$ be arbitrary real numbers. Prove that
$$\sum_{i, j=1}^{n}\left|a_{i}+a_{j}\right| \geq n \sum_{i=1}^{n}\left|a_{i}\right|$$ | Solution. Separating the sequence $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ into two sub-sequences of nonnegative and negative elements
$$\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}=\left\{b_{1}, b_{2}, \ldots, b_{r}\right\} \cup\left\{-c_{1},-c_{2}, \ldots,-c_{s}\right\}$$
with $n=r+s, b_{i} \geq 0 \forall i \in\{1,2,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,379 |
Problem 15. Let $a, b, c$ be non-negative real numbers. Prove that
$$3(a+b+c) \geq 2\left(\sqrt{a^{2}+b c}+\sqrt{b^{2}+c a}+\sqrt{c^{2}+a b}\right)$$ | Solution. WLOG, we may assume that $a \geq b \geq c$. Hence
$$\sqrt{b^{2}+c a}+\sqrt{c^{2}+a b} \leq \sqrt{2\left(b^{2}+c^{2}\right)+2 a(b+c)}$$
We need to prove that
$$2 \sqrt{2\left(b^{2}+c^{2}\right)+2 a(b+c)}+2 \sqrt{a^{2}+b c} \leq 3(a+b+c)$$
Let $s=\frac{1}{2}(b+c)$. Squaring both sides, we obtain an equivalent... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,380 |
Problem 16. Suppose that $a, b, c$ are three non-negative real numbers. Prove that
$$\frac{1}{a^{2}+b^{2}}+\frac{1}{b^{2}+c^{2}}+\frac{1}{c^{2}+a^{2}} \geq \frac{10}{(a+b+c)^{2}}$$ | SOLution. WLOG, assume that $c=\min (a, b, c)$, then
$$\begin{array}{l}
b^{2}+c^{2} \leq\left(b+\frac{c}{2}\right)^{2}=x^{2} \\
a^{2}+c^{2} \leq\left(a+\frac{c}{2}\right)^{2}=y^{2} \\
a^{2}+b^{2} \leq\left(a+\frac{c}{2}\right)^{2}+\left(b+\frac{c}{2}\right)^{2}=x^{2}+y^{2}
\end{array}$$
We deduce that
$$\begin{aligned... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,381 |
Problem 17. Let $a, b, c, d$ be positive real numbers such that $a b c d=1$. Prove that
$$\left(1+a^{2}\right)\left(1+b^{2}\right)\left(1+c^{2}\right)\left(1+d^{2}\right) \geq(a+b+c+d)^{2}$$ | Solution. Because $a b c d=1$, there are two numbers among $a, b, c, d$ that are both not smaller than 1 or not bigger than 1. WLOG, suppose that they are $b$ and $d$, then $(b-1)(d-1) \geq 0$. Applying Cauchy-Schwarz inequality, we get
$$\begin{aligned}
\left(1+a^{2}\right)\left(1+b^{2}\right)\left(1+c^{2}\right)\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,382 |
Problem 18. Let $a, b, c$ be positive real numbers such that $a+b+c=1$. Prove that
$$\frac{a b}{\sqrt{a b+b c}}+\frac{b c}{\sqrt{b c+c a}}+\frac{c a}{\sqrt{c a+a b}} \leq \frac{1}{\sqrt{2}}$$ | Solution. The above inequality is equivalent to
$$\sum_{c y c} a \cdot \sqrt{\frac{b}{a+c}} \leq \frac{1}{\sqrt{2}} \Leftrightarrow \sum_{c y c} \frac{a+b}{2} \cdot \sqrt{\frac{a^{2} b}{(a+c)(a+b)^{2}}} \leq \frac{1}{\sqrt{2}}$$
Using the weighted Jensen inequality for the concave function $f(x)=\sqrt{x}$, we obtain
$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,383 |
Problem 19. Let \(x, y, z\) be non-negative real numbers such that \(x+y+z=1\). Find the maximum of
\[
\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}}
\] | Solution. First we consider the problem in the case $\min (x, y, z)=0$. WLOG, suppose that $z=0$, then $x+y=1$ and therefore
$$\sum_{c y c} \frac{x-y}{\sqrt{x+y}}=\frac{x-y}{\sqrt{x+y}}+\sqrt{y}-\sqrt{x}=x-y+\sqrt{y}-\sqrt{x}=u(v-1),$$
where $u=\sqrt{x}-\sqrt{y}, v=\sqrt{x}+\sqrt{y}$ and $u^{2}+v^{2}=2$.
Denote $u^{2}(... | \sqrt{\frac{71-17 \sqrt{17}}{32}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,384 |
Problem 20. Let \( x, y, z \) be three real numbers in \([-1,1]\) such that \( x + y + z = 0 \).
Prove that
\[
\sqrt{1 + x + y^2} + \sqrt{1 + y + z^2} + \sqrt{1 + z + x^2} \geq 3
\] | Solution. First, we will prove that if $a b \geq 0$ then
$$\sqrt{1+a}+\sqrt{1+b} \geq 1+\sqrt{1+a+b}$$
Indeed, after squaring, the inequality becomes
$$\begin{array}{c}
2+a+b+2 \sqrt{(1+a)(1+b)} \geq 2+a+b+2 \sqrt{1+a+b} \\
\Leftrightarrow(1+a)(1+b) \geq 1+a+b \Leftrightarrow a b \geq 0
\end{array}$$
and we are done. ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,385 |
Problem 21. Suppose that $a, b, c$ are three non-negative real numbers satisfying $ab + bc + ca = 1$. Prove the inequality
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \geq \frac{5}{2}$$ | SOLution. We denote \( x = a + b + c \) and \( z = abc \). The inequality becomes
\[
\begin{array}{c}
2 \sum_{cyc}(a+b)(a+c) \geq 5 \prod_{cyc}(a+b) \Leftrightarrow 6 + 2 \sum_{cyc} a^2 \geq 5(a + b + c - abc) \\
\Leftrightarrow 2x^2 - 5x + 2 + 5z \geq 0 \Leftrightarrow (x-2)(2x-1) + 5z \geq 0
\end{array}
\]
If \( x \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,386 |
Problem 22. Prove the following inequality
$$(\sqrt{2})^{n}\left(a_{1}+a_{2}\right)\left(a_{2}+a_{3}\right) \ldots\left(a_{n}+a_{1}\right) \leq\left(a_{1}+a_{2}+a_{3}\right)\left(a_{2}+a_{3}+a_{4}\right) \ldots\left(a_{n}+a_{1}+a_{2}\right)$$
where \(a_{1}, a_{2}, \ldots, a_{n}\) are arbitrary positive real numbers. | Solution. According to the following results
$$\begin{aligned}
\left(a_{1}+a_{2}+a_{3}\right)^{2} & \geq\left(2 a_{1}+a_{2}\right)\left(a_{2}+2 a_{3}\right) \\
\left(2 a_{1}+a_{2}\right)\left(2 a_{2}+a_{1}\right) & =2 a_{1}^{2}+2 a_{2}^{2}+5 a_{1} a_{2} \geq 2\left(a_{1}+a_{2}\right)^{2}
\end{aligned}$$
we are done imm... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,387 |
Example 1.2.1. Let $a, b, c$ be positive real numbers with sum 3. Prove that
$$\frac{a}{1+b^{2}}+\frac{b}{1+c^{2}}+\frac{c}{1+a^{2}} \geq \frac{3}{2}$$ | Solution. In fact, it's impossible to use AM-GM for the denominators because the sign will be reversed
$$\frac{a}{1+b^{2}}+\frac{b}{1+c^{2}}+\frac{c}{1+a^{2}} \leq \frac{a}{2 b}+\frac{b}{2 c}+\frac{c}{2 a} \geq \frac{3}{2} ?!$$
However, we can use the same application in another appearance
$$\frac{a}{1+b^{2}}=a-\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,388 |
Problem 23. Let $a, b, c$ be non-negative real numbers with sum 3. Prove that
$$\frac{a b+b c+c a}{a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}} \geq \frac{a^{3}+b^{3}+c^{3}}{36}$$ | Solution. WLOG, we may assume that \(a \geq b \geq c\). Denote
\[ f(a, b, c) = 36(ab + bc + ca) - (a^3 + b^3 + c^3)(a^3 b^3 + b^3 c^3 + c^3 a^3). \]
We will prove \(f(a, b, c) \geq f(a, b+c, 0)\). Indeed,
\[ \begin{array}{l}
a^3 + b^3 + c^3 \leq a^3 + (b+c)^3 ; a^3 b^3 + b^3 c^3 + c^3 a^3 \leq a^3 (b+c)^3 \\
\Rightarr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,389 |
Problem 24. Let \(a, b, c, d\) be four real numbers satisfying that \(\left(1+a^{2}\right)\left(1+b^{2}\right)(1+\) \(\left.c^{2}\right)\left(1+d^{2}\right)=16\). Prove the inequality
\[
-3 \leq a b + b c + c a + d a + a c + b d - a b c d \leq 5
\] | Solution. We denote $S=a b+b c+c d+d a+a c+b d-a b c d$, then
$$S-1=(1-a b)(c d-1)+(a+b)(c+d)$$
Applying Cauchy-Schwarz inequality, we obtain
$$\begin{aligned}
(S-1)^{2} & \leq\left((1-a b)^{2}+(a+b)^{2}\right)\left((1-c d)^{2}+(c+d)^{2}\right) \\
& =\left(1+a^{2}\right)\left(1+b^{2}\right)\left(1+c^{2}\right)\left(1+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,390 |
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