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x - 1/y = y - 1/z = z - 1/x (German MO)
aether_1729   2
N 15 minutes ago by rchokler
Find all real $x, y, z$ such that
\[x - \frac{1}{y} = y - \frac{1}{z} = z - \frac{1}{x}.\]
Source: Problem 520943 from 52nd German MO, final round / day 1, grade 9.
2 replies
aether_1729
Yesterday at 12:45 PM
rchokler
15 minutes ago
Inequalities
sqing   5
N 2 hours ago by sqing
Let $ a,b\geq1 $ and $ a+b+ab=8 .$ Prove that
$$\frac{3\sqrt{5}}{2}\geq \frac{\sqrt{a^2-1}}{b^2}+\frac{\sqrt{b^2-1}}{a^2}\geq \frac{\sqrt{3}}{2} $$
5 replies
sqing
Yesterday at 7:36 AM
sqing
2 hours ago
No more topics!
Inequalities
sqing   55
N Nov 21, 2024 by sqing
Let $a,b>0$ and $ab^2(a+b)=9.$ Prove that
$$2a+5b\geq 2\sqrt[4]{27(3+8\sqrt{6})} $$$$2a+9b\geq 6\sqrt[4]{48\sqrt{2}-39} $$
55 replies
sqing
Oct 31, 2024
sqing
Nov 21, 2024
Inequalities
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ytChen
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#42
Y by
sqing wrote:
Let $ a,b,c\geq 0 .$ Prove that$$ \frac{a}{b+c}+ \frac{b }{c+a}+ \frac{4c}{a+b+c}\geq 2$$

Solution. Cauchy-Schwartz’s inequality shows
\begin{align*}& \frac{a}{b+c}+ \frac{b }{c+a}+ \frac{4c}{a+b+c}\\
=&\frac{a+b+c}{b+c}-1+ \frac{a+b+c}{c+a}-1+ \frac{4c}{a+b+c}\\
\ge&\frac{4(a+b+c)}{a+b+2c}+ \frac{4c}{a+b+2c}-2=4-2=2,
\end{align*}and the minimum value $2$ can be attained if $a=b$ and $c=0$. $\blacksquare$
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ytChen
1052 posts
#43
Y by
sqing wrote:
Let $a,b,c >0 .$ Prove that $$ \frac{a^3}{b+1}+ \frac{b^3}{c+1}+ \frac{c^3}{a+1}\geq \frac{3abc}{abc+1}$$
Remark. When $0<a=b=c<1$, we get
\begin{align*} & \frac{a^3}{b+1}+ \frac{b^3}{c+1}+ \frac{c^3}{a+1}=\frac{3a^3}{a+1}< \frac{3a^3}{a^3+1}= \frac{3abc}{abc+1}.
\end{align*}Hence the result should be wrong.
This post has been edited 1 time. Last edited by ytChen, Nov 16, 2024, 7:53 PM
Reason: Typo
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ytChen
1052 posts
#44
Y by
sqing wrote:
Let $a,b,c >0 .$ Prove that
$$   \frac{ab}{(a+b)^2+c^2}+\frac{bc}{(b+c)^2+a^2}+ \frac{ca}{(c+a)^2+b^2} \leq \frac{3}{5}$$
Solution. Cauchy-Schwartz’s inequality shows
\begin{align*}& \frac{ab}{(a+b)^2+c^2}+\frac{bc}{(b+c)^2+a^2}+ \frac{ca}{(c+a)^2+b^2}\\
\le&\frac{ab}{4ab+c^2}+\frac{bc}{4bc+a^2}+ \frac{ca}{4ca+b^2}\\
=&\frac{1}{4}\left[3-\frac{c^2}{4ab+c^2}-\frac{a^2}{4bc+a^2}- \frac{b^2}{4ca+b^2}\right]\\
\le& \frac{1}{4}\left[3-\frac{(a+b+c)^2}{2(ab+bc+ca)+(a+b+c)^2}\right]\\
=& \frac{1}{4}\left[3-\frac{1}{\frac{2(ab+bc+ca)}{(a+b+c)^2 }+1}\right]\le\frac{1}{4}\left[3-\frac{1}{\frac{2}{3}+1}\right] =\frac{3}{5},
\end{align*}and the maximum value $\frac{3}{5}$ can be attained if $a=b=c$. $\blacksquare$
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#45
Y by
Sorry .Thank ytChen .

Let $a,b,c >0 .$ Prove that $$ \frac{a^3}{b^3+1}+ \frac{b^3}{c^3+1}+ \frac{c^3}{a^3+1}\geq \frac{3abc}{abc+1}$$
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sqing
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#46
Y by
Very very nice .Thank ytChen .
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ytChen
1052 posts
#47
Y by
sqing wrote:
Let $a,b,c >0 .$ Prove that $$ \frac{a^3}{b^3+1}+ \frac{b^3}{c^3+1}+ \frac{c^3}{a^3+1}\geq \frac{3abc}{abc+1}$$
Solution. First recall inequality $$(x+y+z)^2\ge3(xy+yz+zy).$$Thus, by Cauchy-Schwartz’s inequality, we get
\begin{align*}& \frac{a^3}{b^3+1}+ \frac{b^3}{c^3+1}+ \frac{c^3}{a^3+1}=\\
& \frac{a^6}{a^3\left(b^3+1\right)}+ \frac{b^6}{b^3\left(c^3+1\right)}+ \frac{c^6}{c^3\left(a^3+1\right)}\\
\ge& \frac{\left(a^3+b^3+c^3\right)^2}{a^3b^3+b^3c^3+c^3a^3+ a^3+b^3+c^3}\\
=& \frac{1}{\frac{a^3b^3+b^3c^3+ c^3a^3}{\left(a^3+b^3+c^3\right)^2 }+\frac{1}{a^3+b^3+c^3}}\\
\ge&\frac{1}{\frac{1}{3}+\frac{1}{3abc}}=\frac{3abc}{abc+1},
\end{align*}and the equality occurs if $a=b=c=1$. $\blacksquare$
This post has been edited 2 times. Last edited by ytChen, Nov 20, 2024, 11:52 PM
Reason: Typo
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sqing
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#48
Y by
Very very nice .Thank ytChen .
Let $ a,b,c\ge 0 $ and $   ab+bc+ca+abc=4 $.Prove that
$$(a+b+c-4)^2+4 \ge 5abc $$Equality holds when $(a,b,c)=\left(\frac{2}{3}, \frac{2}{3},2\right).$
This post has been edited 1 time. Last edited by sqing, Nov 21, 2024, 11:24 AM
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ytChen
1052 posts
#49
Y by
sqing wrote:
Let $a,b,c >0 .$ Prove that
$$ \frac{ab}{(a+b)^2+c^2}+\frac{bc}{(b+c)^2+a^2}+ \frac{ca}{(c+a)^2+b^2} \leq \frac{3}{5}$$

Solution.Cauchy-Schwartz’s inequality shows
\begin{align*}& \frac{ab}{(a+b)^2+c^2}+\frac{bc}{(b+c)^2+a^2}+ \frac{ca}{(c+a)^2+b^2}\\
\le& \frac{ab}{4ab+c^2}+\frac{bc}{4bc+a^2}+ \frac{ca}{4ca+b^2}\\
=& \frac{1}{4}\left[3-\frac{c^2}{4ab+c^2}-\frac{a^2}{4bc+a^2}- \frac{b^2}{4ca+b^2}\right]\\
\le& \frac{1}{4}\left[3-\frac{(a+b+c)^2}{(a+b+c)^2+2(ab+bc+ac)}\right]\\
=& \frac{1}{4}\left[3-\frac{1}{1+\frac{2(ab+bc+ca)}{(a+b+c)^2}}\right]\le \frac{1}{4}\left[3-\frac{1}{1+\frac{2}{3}}\right] =\frac{3}{5},
\end{align*}and the maximum value $\frac{3}{5} $ can be attained if $a=b=c$. $\blacksquare$
This post has been edited 2 times. Last edited by ytChen, Nov 21, 2024, 1:49 AM
Reason: Typo
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sqing
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#50
Y by
Very very nice .Thank ytChen .
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sqing
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#51
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Let $ a,b,c\geq 0 .$ Prove that$$ \frac{a}{b+c}+ \frac{b }{c+a}+ \frac{ c+\sqrt{ab} }{a+b}\geq 2$$
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ytChen
1052 posts
#52
Y by
sqing wrote:
Let $ a,b,c\geq 0 .$ Prove that$$ \frac{a}{b+c}+ \frac{b }{c+a}+ \frac{ c+\sqrt{ab} }{a+b}\geq 2$$
Solution. Without loss of generality, assume $a\ge b\ge0$. For $\frac{a+b }{c+\sqrt{ab}}+ \frac{ c+\sqrt{ab} }{a+b}\ge2$, it suffices to show
$$\frac{a}{b+c}+ \frac{b }{c+a}-\frac{a+b}{c+\sqrt{ab} }\ge0.$$Indeed,
\begin{align*}& \frac{a}{b+c}+ \frac{b }{c+a}-\frac{a+b}{c+\sqrt{ab} }\\
=& \frac{a}{b+c}-\frac{a}{c+\sqrt{ab} } + \frac{b }{c+a}-\frac{b}{c+\sqrt{ab} }\\
=& \frac{a\sqrt b\left(\sqrt a-\sqrt b\right)}{(b+c)\left(c+\sqrt{ab}\right)}+ \frac{b\sqrt a\left(\sqrt b-\sqrt a\right)}{(c+a)\left(c+\sqrt{ab}\right)}\\
\ge& \frac{a\sqrt b\left(\sqrt a-\sqrt b\right)}{(a+c)\left(c+\sqrt{ab}\right)}+ \frac{b\sqrt a\left(\sqrt b-\sqrt a\right)}{(c+a)\left(c+\sqrt{ab}\right)}\\
=& \frac{\sqrt{ab}\left(\sqrt a-\sqrt b\right)^2}{(a+c)\left(c+\sqrt{ab}\right)}\ge0,
\end{align*}and the equality occurs if $b=0$ and $a=c$. $\blacksquare$
This post has been edited 2 times. Last edited by ytChen, Nov 21, 2024, 7:48 AM
Reason: Typo
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sqing
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#53
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Very very nice.Thank ytChen:
Let $ a,b,c > 0 .$ Then$$\frac{a}{b+c}+ \frac{b }{a+c} \geq\frac{a+b}{\sqrt{ab}+c}$$h
This post has been edited 3 times. Last edited by sqing, Nov 21, 2024, 8:22 AM
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#54
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Let $ a,b > 0 .$ Prove that $$\frac{a^2}{b+1}+ \frac{b^2 }{a+1} \geq\frac{a^2+b^2}{\sqrt{ab} +1}$$
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ytChen
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#55
Y by
sqing wrote:
Let $ a,b,c\geq 0 .$ Prove that$$ \frac{a}{b+c}+ \frac{b }{c+a}+ \frac{4(c+\sqrt{ab})}{a+b}\geq 4$$

Solution. Without loss of generality, assume $a\ge b\ge0$. Since $$\frac{a+b }{c+\sqrt{ab}}+ \frac{4\left(c+\sqrt{ab}\right)}{a+b}\ge4,$$it suffices to show
$$\frac{a}{b+c}+ \frac{b }{c+a}-\frac{a+b}{c+\sqrt{ab} }\ge0,$$which was proved in post #53, and the equality occurs if $b=0$ and $a=2c$. $\blacksquare$
This post has been edited 1 time. Last edited by ytChen, Nov 21, 2024, 8:43 AM
Reason: Typo
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#56
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Very very nice.Thank ytChen.
This post has been edited 1 time. Last edited by sqing, Nov 21, 2024, 9:06 AM
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