# Math Help - inequalty

1. ## inequalty

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

prove

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

2. ## 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

3. ## Re: inequalty

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.