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Source: Iran TST 2015, second exam, day 2, problem 3

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164 posts

#1 h May 17, 2015, 3:49 PM • 3 Y

Y by quykhtn-qa1, mathbook, Adventure10

If $a,b,c$ are positive real numbers such that $a+b+c=abc$ prove that
$\frac{abc}{3\sqrt{2}}\left ( \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \right )\geq \sum_{cyc}\frac{a}{a^2+1}$

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2105 posts

#2 h May 17, 2015, 4:42 PM • 5 Y

Y by dm_edogawasonichi, mathbook, Adventure10, Mango247, MS_asdfgzxcvb

Stronger:

If $a,b,c$ are positive real numbers such that $a+b+c=abc$ prove that
$\frac{abc}{3\sqrt{2}\sqrt[4]{27}}\left ( \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \right )\geq \sum_{cyc}\frac{a}{a^2+1}$

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1347 posts

#3 h May 18, 2015, 3:06 AM • 11 Y

Y by dm_edogawasonichi, mathbook, Dadgarnia, A_Gappus, baladin, Gaussian_cyber, Nathanisme, DanDumitrescu, Adventure10, Mango247, MS_asdfgzxcvb

Dadgarnia wrote:

If $a,b,c$ are positive real numbers such that $a+b+c=abc$ prove that
$\frac{abc}{3\sqrt{2}}\left ( \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \right )\geq \sum_{cyc}\frac{a}{a^2+1}$

I think this inequality is not a nice problem in Iran TST. We can prove the stronger inequality of socrates by AM-GM:
$\frac{abc}{3\sqrt{2}\sqrt[4]{27}}\left ( \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \right )\geq \sum_{cyc}\frac{a}{a^2+1}$
Indeed, since $\begin{aligned} \dfrac{1}{abc} \cdot \sum_{cyc} \dfrac{a}{a^2+1} & =\sum_{cyc}\dfrac{1}{a^2bc+bc}=\sum_{cyc}\dfrac{1}{a(a+b+c)+bc} =\sum_{cyc}\dfrac{1}{(a+b)(a+c)} \\ & =\dfrac{2(a+b+c)}{(a+b)(b+c)(c+a)} \leq \dfrac{2abc}{8abc}=\dfrac{1}{4}. \end{aligned}$ It suffices to show that $\sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \geq \dfrac{3\sqrt{2}\sqrt[4]{27}}{4}.$ Notice that
$(a+b)(b+c)(c+a) \geq \dfrac{8(a+b+c)(ab+bc+ca)}{9} \geq \dfrac{8(a+b+c)\sqrt{3abc(a+b+c)}}{9}=\dfrac{8\sqrt{3} a^2b^2c^2}{9}.$ Hence,
$(a^3+b^3)(b^3+c^3)(c^3+a^3) \geq ab(a+b)\cdot bc(b+c) \cdot ca(c+a) \geq \dfrac{8\sqrt{3}a^4b^4c^4}{9}.$ Now, using the AM-GM inequality, we have $(ab+1)(bc+1)(ca+1)=\dfrac{(abc+a)(abc+b)(abc+c) }{abc}\leq \dfrac{1}{abc}\left(\dfrac{abc+a+abc+b+abc+c}{3}\right)^3=\dfrac{64a^2b^2c^2}{27}.$ Therefore, $\sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \geq 3\sqrt[3]{\dfrac{\sqrt{(a^3+b^3)(b^3+c^3)(c^3+a^3)}}{(ab+1)(bc+1)(ca+1)}} \geq 3\sqrt[3]{\dfrac{\sqrt{\dfrac{8\sqrt{3} a^4b^4c^4}{9}}}{\dfrac{64a^4b^4c^4}{27}}}=\dfrac{3\sqrt{2}\sqrt[4]{27}}{4}.$ The proof is completed. Equality holds for $a=b=c=\sqrt{3}.$

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41348 posts

sqing

#4 h May 18, 2015, 4:19 AM • 3 Y

Y by mathbook, Adventure10, Mango247

The stronger inequality
If $a,b,c$ are positive real numbers such that $a+b+c=abc$ prove that $\frac{1}{\sqrt[4]{12}}\left ( \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \right )\geq \sum_{cyc}\frac{a}{a^2+1}.$

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1347 posts

#6 h May 18, 2015, 5:35 AM • 3 Y

Y by dm_edogawasonichi, Wangzu, Adventure10

sqing wrote:

The stronger inequality
If $a,b,c$ are positive real numbers such that $a+b+c=abc$ prove that $\frac{1}{\sqrt[4]{12}}\left ( \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \right )\geq \sum_{cyc}\frac{a}{a^2+1}.$

It follows from
$\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}+\dfrac{c}{c^2+1}=\dfrac{2a^2b^2c^2}{(a+b)(b+c)(c+a)} \leq \dfrac{2a^2b^2c^2}{\dfrac{8\sqrt{3} a^2b^2c^2}{9}}=\dfrac{9}{4\sqrt{3}}.$

$\sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \geq \dfrac{3\sqrt{2}\sqrt[4]{27}}{4}.$

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1135 posts

#7 h Jun 12, 2015, 9:34 PM • 2 Y

Y by Adventure10, Mango247

quykhtn-qa1 wrote:

Dadgarnia wrote:

If $a,b,c$ are positive real numbers such that $a+b+c=abc$ prove that
$\frac{abc}{3\sqrt{2}}\left ( \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \right )\geq \sum_{cyc}\frac{a}{a^2+1}$

I think this inequality is not a nice problem in Iran TST. We can prove the stronger inequality of socrates by AM-GM:
$\frac{abc}{3\sqrt{2}\sqrt[4]{27}}\left ( \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \right )\geq \sum_{cyc}\frac{a}{a^2+1}$
Indeed, since $\begin{aligned} \dfrac{1}{abc} \cdot \sum_{cyc} \dfrac{a}{a^2+1} & =\sum_{cyc}\dfrac{1}{a^2bc+bc}=\sum_{cyc}\dfrac{1}{a(a+b+c)+bc} =\sum_{cyc}\dfrac{1}{(a+b)(a+c)} \\ & =\dfrac{2(a+b+c)}{(a+b)(b+c)(c+a)} \leq \dfrac{2abc}{8abc}=\dfrac{1}{4}. \end{aligned}$ It suffices to show that $\sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \geq \dfrac{3\sqrt{2}\sqrt[4]{27}}{4}.$ Notice that
$(a+b)(b+c)(c+a) \geq \dfrac{8(a+b+c)(ab+bc+ca)}{9} \geq \dfrac{8(a+b+c)\sqrt{3abc(a+b+c)}}{9}=\dfrac{8\sqrt{3} a^2b^2c^2}{9}.$ Hence,
$(a^3+b^3)(b^3+c^3)(c^3+a^3) \geq ab(a+b)\cdot bc(b+c) \cdot ca(c+a) \geq \dfrac{8\sqrt{3}a^4b^4c^4}{9}.$ Now, using the AM-GM inequality, we have $(ab+1)(bc+1)(ca+1)=\dfrac{(abc+a)(abc+b)(abc+c) }{abc}\leq \dfrac{1}{abc}\left(\dfrac{abc+a+abc+b+abc+c}{3}\right)^3=\dfrac{64a^2b^2c^2}{27}.$ Therefore, $\sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \geq 3\sqrt[3]{\dfrac{\sqrt{(a^3+b^3)(b^3+c^3)(c^3+a^3)}}{(ab+1)(bc+1)(ca+1)}} \geq 3\sqrt[3]{\dfrac{\sqrt{\dfrac{8\sqrt{3} a^4b^4c^4}{9}}}{\dfrac{64a^4b^4c^4}{27}}}=\dfrac{3\sqrt{2}\sqrt[4]{27}}{4}.$ The proof is completed. Equality holds for $a=b=c=\sqrt{3}.$

Your proof is truly amazing! What would be some motivations for this solution? Thanks!

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42 posts

GaU1Er

#8 h Oct 9, 2021, 7:48 PM

Y by

After homogenizing the given inequality is equivalent to :
$\sqrt[4]{A}(\sum{\frac{\sqrt{a^3+b^3}}{ab(A+c)}})$ $\ge$ $\frac{6\sqrt{2}\sqrt[4]{(abc)^3}}{(a+b)(b+c)(c+a)}$ , where $A=a+b+c$
Now by AM-GM we have that : $LHS$ $\ge$ $\frac{3\sqrt{2}\sqrt[4]{27(abc)^3}}{\sqrt[3]{(abc)^2(A+a)(A+b)(A+c)}}$
Thus we must show that : $\frac{\sqrt[4]{27}}{\sqrt[3]{(abc)^2 \prod(A+a)}}$ $\ge$ $\frac{2}{(a+b)(b+c)(c+a)}$
But we'll prove the stronger inequality : $(a+b)(b+c)(c+a)$ $\ge$ $2\sqrt[3]{(abc)^2\prod{(A+a)}}$
Again using AM-GM $LHS=\frac{\sum{(Aa+bc)(b+c)}}{3}$ $\ge$ $2\sqrt[3]{(abc)\prod{(Aa+bc)}}$
So it suffices to show that $\prod{(Aa+bc)}$ $\ge$ $(abc) \prod{(A+a)}$
After expending it remains to prove that $A^2((ab)^2+(bc)^2+(ac)^2)+Aabc(a^2+b^2+c^2)$ $\ge$ $A^2(abc)(a+b+c)+A(abc)(ab+bc+ac)$
wich is obviously true .

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671 posts

#9 h Mar 27, 2025, 11:16 AM • 1 Y

Y by L13832

We will prove the stronger inequality below.
$\frac{abc}{3\sqrt{2}\sqrt[4]{27}}\left ( \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \right ) \overset{?}{\geq} \sum_{cyc}\frac{a}{a^2+1}$ Let $\frac{1}{bc}=x^2,\frac{1}{ac}=y^2,\frac{1}{ab}=z^2$ so $a=\frac{x}{yz},b=\frac{y}{xz},c=\frac{z}{xy}$ and $x^2+y^2+z^2=1$ .
$\frac{1}{xyz. 3\sqrt2\sqrt[4]{27}}(\sum{\frac{\sqrt[3]{\frac{y^3}{x^3z^3}+\frac{z^3}{x^3y^3}}}{\frac{1}{x^2}+1}})\overset{?}{\geq} \sum{\frac{\frac{x}{yz}}{\frac{x^2}{y^2z^2}+1}}\iff \frac{1}{3\sqrt2\sqrt[4]{27}}(\sum{\frac{\sqrt{\frac{xy^3}{z^3}+\frac{xz^3}{y^3}}}{x^2+1}})\overset{?}{\geq} x^2y^2z^2\sum{\frac{1}{x^2+y^2z^2}}$ $\frac{1}{3\sqrt 2\sqrt[4]{27}}\sum{\sqrt{\frac{x(\frac{y^3}{z^3}+\frac{z^3}{y^3})}{x^2+1}}}\geq \frac{1}{3\sqrt[4]{27}}\sum{\frac{\sqrt{x}}{x^2+1}}\overset{C.S}{\geq} \frac{1}{3\sqrt[4]{27}}.\frac{(\sum{\sqrt[4] x})^2}{4}\geq \frac{3\sqrt[6]{xyz}}{4\sqrt[4]{27}}$ $x^2y^2z^2\sum{\frac{1}{\frac{x^2}{3}+\frac{x^2}{3}+\frac{x^2}{3}+y^2z^2}}\leq x^2y^2z^2\sum{\frac{1}{\frac{4x\sqrt{xyz}}{\sqrt[4]{27}}}}=\frac{x^2y^2z^2\sqrt[4]{27}}{4\sqrt{xyz}}.\sum{\frac{1}{x}}\leq \frac{(xyz)^{\frac{3}{2}}\sqrt[4]{27}}{4}.\frac{1}{xyz}=\frac{\sqrt{xyz}.\sqrt[4]{27}}{4}$ And $\frac{\sqrt{xyz}.\sqrt[4]{27}}{4}\leq \frac{3\sqrt[6]{xyz}}{4\sqrt[4]{27}}\iff \sqrt[3]{xyz}\overset{?}{\leq} \frac{1}{\sqrt3}\iff xyz\overset{?}{\leq} \frac{1}{3\sqrt 3}$ which is true by AM-GM as desired. $\blacksquare$

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