Inequality Problem

Suppose $a,b,c$ be 3 positive reals such that

$a+b+c≥\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$

Show that $\frac{a^{3}c}{b(c+a)}+\frac{b^{3}a}{c(a+b)}+\frac{c^{3}b}{a(b+c)}≥\frac{3}{2}$ .

I cannot still solve this problem.Please,someone help me .

#Algebra #CauchySchwarzInequality #Inequality

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

$\displaystyle 2(a+b+c)\left( \frac{a}{b} +\frac{b}{c} +\frac{c}{a} \right) \left( \sum_{\text{cyc}}\frac{a^3c}{b(c+a)} \right)$

$\displaystyle \stackrel{\text{Hölder}}\ge (a+b+c)^3\stackrel{\text{?}}\ge 3(a+b+c)\left( \frac{a}{b} +\frac{b}{c} +\frac{c}{a} \right)$

$\displaystyle 3\left( \frac{a}{b} +\frac{b}{c} +\frac{c}{a} \right)\le 3(a+b+c)\stackrel{\text{?}}\le (a+b+c)^2$

$\displaystyle a+b+c\ge \frac{a}{b} +\frac{b}{c} +\frac{c}{a}\stackrel{\text{AM-GM}}\ge 3.\:\:\square$

mathh mathh - 6 years, 10 months ago

i think you should try to use AM-GM inequality

akash deep - 6 years, 10 months ago

I have tried...but it was of no use

Souryajit Roy - 6 years, 10 months ago

search up holder's inequality

blah blah - 6 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$