if $\displaystyle a,b,c > 0$ and $\displaystyle a+b+c=1$

prove

$\displaystyle \frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3} {c^2+a^2}\geq\frac{1}{ 2}$

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- Mar 7th 2008, 12:10 PMperashinequalty
if $\displaystyle a,b,c > 0$ and $\displaystyle a+b+c=1$

prove

$\displaystyle \frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3} {c^2+a^2}\geq\frac{1}{ 2}$ - Sep 16th 2013, 10:13 PMbrosnan123Re: inequalty
Solution:

Given a,b,c>0

a+b+c=1

this is possible when a=b=c=1/3

a^3/(a²+b²)+b^3/(b²+c²)+c^3/(c²+a²)>=1/2

then (1/3)²/(1/9+1/9)+ (1/3)²/(1/9+1/9)+(1/3)²/(1/9+1/9)

=1/2

hence proved - Sep 17th 2013, 03:14 AMShakarriRe: inequalty