easy inequality with a^2+b^2+c^2=3abc

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easy inequality with a^2+b^2+c^2=3abc

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Sayan

#1 h Jun 29, 2012, 3:43 AM • 3 Y

Y by integrated_JRC, Adventure10, Mango247

Let $a,b,c$ be positive reals such that $a^2+b^2+c^2=3abc$ . Prove that
$\frac{a}{b^2c^2}+\frac{b}{c^2a^2}+\frac{c}{a^2b^2} \geq \frac{9}{a+b+c}$

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3428 posts

#2 h Jun 29, 2012, 5:15 AM • 4 Y

Y by ayan_mathematics_king, Adventure10, Mango247, ehuseyinyigit

Sayan wrote:

Let $a,b,c$ be positive reals such that $a^2+b^2+c^2=3abc$ . Prove that
$\frac{a}{b^2c^2}+\frac{b}{c^2a^2}+\frac{c}{a^2b^2} \geq \frac{9}{a+b+c}$

$LHS=\frac{a^4}{a^3b^2c^2}+\frac{b^4}{b^3c^2a^2}+\frac{c^4}{c^3a^2b^2}\ge\frac{(a^2+b^2+c^2)^2}{a^2b^2c^2(a+b+c)}=\frac{9}{a+b+c}$
I used C-S and $a^2+b^2+c^2=3abc$

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555 posts

#3 h Jun 29, 2012, 5:18 AM • 4 Y

Y by RISH2909, Adventure10, Mango247, and 1 other user

Probably the same but it just looks nicer...

Divide by $abc$ the condition and we get $\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}=3$

By C-S $(\frac{a}{b^2c^2}+\frac{b}{a^2c^2}+\frac{c}{a^2b^2}) (a+b+c) \ge (\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab})^2 = 9$ and we are done.

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#4 h Jun 29, 2012, 5:19 AM • 2 Y

Y by Adventure10, Mango247

we can change it to
$(9.a^3/(a^2+b^2+c^2)^2)+(9.b^3/(a^2+b^2+c^2)^2)+(9.(c^3/a^2+b^2+c^2)^2)>=(9/a+b+c)$ so we have
$a^3+b^3+c^3/(a^2+b^2+c^2)^2>=1/a+b+c$
$(a^3+b^3+c^3).(a+b+c)>=(a^2+b^2+c^2)^2$
and its a famous inequality

This post has been edited 1 time. Last edited by sahadian, Jan 22, 2013, 11:06 AM

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#5 h Jun 29, 2012, 5:20 AM • 2 Y

Y by Adventure10, Mango247

i think these all answer are like each other

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#6 h Jun 29, 2012, 8:20 AM • 2 Y

Y by Adventure10, Mango247

we have
$\frac{a^{3}+b^{3}+c^{3}}{a^{2}b^{2}c^{2}}\geq \frac{9}{a+b+c}\Leftrightarrow (\sum a)(\sum a^{3})\geq (\sum a^{2})^{2}$

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#7 h Aug 13, 2012, 5:54 PM • 2 Y

Y by Adventure10, Mango247

$\sum \frac{a}{b^2.c^2}$ = $\sum\frac{a^4}{a^3.b^2.c^2} \ge \frac{[\sum a^2]^2}{a^2.b^2.c^2(a+b+c)}$ = $\frac{9}{a+b+c}$

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676 posts

#8 h Aug 22, 2012, 6:56 AM • 2 Y

Y by Adventure10, Mango247

Sayan wrote:

Let $a,b,c$ be positive reals such that $a^2+b^2+c^2=3abc$ . Prove that
$\frac{a}{b^2c^2}+\frac{b}{c^2a^2}+\frac{c}{a^2b^2} \geq \frac{9}{a+b+c}$

$\frac{a}{b^2c^2}+\frac{b}{c^2a^2}+\frac{c}{a^2b^2} = \frac{a^4}{a^3b^2c^2}+\frac{b^4}{b^3c^2a^2}+\frac{c^3}{c^3a^2b^2} \geq_ \frac{(a^2+b^2+c^2)^2}{a^3b^2c^2 + b^3c^2a^2 + c^3a^2b^2} = \frac{(a^2+b^2+c^2)^2}{a^2b^2c^2(a+b+c)}$

since $a^2+b^2+c^2 = 3abc$

$\frac{(a^2+b^2+c^2)^2}{a^2b^2c^2(a+b+c)} = \frac{(3abc)^2}{a^2b^2c^2(a+b+c)} = \frac{9}{a+b+c}$

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sayantanchakraborty

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sayantanchakraborty

#9 h Oct 14, 2013, 5:29 AM • 2 Y

Y by Adventure10, Mango247

Easy but nice problem. The inequality transforms to $(a^3+b^3+c^3)(a+b+c)>=9a^2b^2c^2$ . Use cauchy scwarz inequality.

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6107 posts

#10 h Nov 13, 2019, 1:25 PM • 2 Y

Y by Adventure10, Mango247

Clearing the denominators, we need to prove
$(a+b+c)(a^3+b^3+c^3)\geq 9a^2b^2c^2$ which is true by C-S

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