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?

$$\dfrac{\sum a^4}2-\left(\dfrac{\sum a^2}2\right)^2=\dfrac{\sum a^4-\sum2b^2c^2}4$$
$$\sum a^4-\sum2b^2c^2=(a^2+b^2-c^2)^2-(2ab)^2$$
$$(a^2+b^2-c^2)^2-(2ab)^2=(a^2+b^2-c^2-2ab)(a^2+b^2-c^2+2ab)$$
$$a^2+b^2-c^2+2ab=(a+b)^2-c^2=(a+b+c)(a+b-c)$$
