Prove an equality If $a+b+c=0$
prove that
$\frac {(a^4 +b^4 +c^4)}{2}=\frac {(a^2+b^2+c^2)}{2^2}^2$
I have expanded the right side and have got this far:
$a^4+b^4+c^4+2(a^2b^2+a^2c^2+b^2c^2)$
I need $a^2=b^2=c^2$ to prove the equality. 
Any ideas?

One has $$0 = (a+b+c)^2 = a^2 + b^2 +c^2 + 2(ab+bc+ca).$$
Then, $$(\frac{a^2 +b^2 +c^2}{2})^2 = (ab+bc+ca)^2 = a^2b^2 + b^2c^2 +c^2a^2 + 2abc(a+b+c) = a^2b^2 + b^2c^2 +c^2a^2.$$
Finally, one has
$$(a^2+b^2+c^2)^2 = a^4+b^4+c^4 + 2(a^2b^2 + b^2c^2 +c^2a^2) = a^4+b^4+c^4 + 2(\frac{a^2 +b^2 +c^2}{2})^2$$
Thus, $$\frac{1}{2}(a^4+b^4+c^4) = (\frac{a^2 +b^2 +c^2}{2})^2$$
