# inequalty

• March 7th 2008, 12:10 PM
perash
inequalty
if $a,b,c > 0$ and $a+b+c=1$

prove

$\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3} {c^2+a^2}\geq\frac{1}{ 2}$
• September 16th 2013, 10:13 PM
brosnan123
Re: 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
• September 17th 2013, 03:14 AM
Shakarri
Re: inequalty
Quote:

Originally Posted by brosnan123
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

This is not a proof, you only showed it was true for one set of values for a, b and c

Edit: I didn't realise this question was from 2008.