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?
 A: Following is a more systematic approach uses Newton's identities.
Let $e_1, e_2, e_3$ be the elementary symmetric polynomials associated with $a, b, c,$ i.e.
$$\begin{cases}
e_1 &= a + b + c\\
e_2 &= ab + bc + ca\\
e_3 &= abc
\end{cases}
\quad\iff\quad  (x-a)(x-b)(x-c) = x^3 - e_1 x^2 + e_2 x - e_3
$$
and let $p_k = a^k + b^k + c^k, k \in \mathbb{Z}_{+}$ be the corresponding power sums.
We are given $e_1 = a + b + c = 0$. Newton identities tell us
$$\require{cancel}
\newcommand{\xxx}[1]{\color{red}{\cancelto{0}{\color{gray}{#1}}}}
\begin{array}{rlclr}
p_1 -\xxx{e_1} &= 0 &\implies& p_1 = 0 \\
p_2 -\xxx{e_1} p_1 + 2 e_2 &= 0 &\implies& p_2 = -2e_2 & (*1a) \\
p_3 -\xxx{e_1} p_2 + e_2 \xxx{p_1} - 3 e_3 &= 0 &\implies& p_3 = 3 e_3\\
p_4 -\xxx{e_1} p_3 + e_2 p_2 - e_3\xxx{p_1} &= 0 &\implies& p_4 = -e_2 p_2 & (*1b)
\end{array}
$$
Combine $(*1a)$ and $(*1b)$, we have
$$a^4 + b^4 + c^4 = p_4 = -e_2 p_2 = \frac12 p_2^2 = \frac12 (a^2 + b^2 + c^2)^2\tag{*2}$$
A: $$\dfrac{\sum a^4}2-\left(\dfrac{\sum a^2}2\right)^2=\dfrac{\sum a^4-\sum2b^2c^2}4$$
$$\sum a^4-\sum2b^2c^2=(a^2+b^2-c^2)^2-(2ab)^2$$
$$(a^2+b^2-c^2)^2-(2ab)^2=(a^2+b^2-c^2-2ab)(a^2+b^2-c^2+2ab)$$
$$a^2+b^2-c^2+2ab=(a+b)^2-c^2=(a+b+c)(a+b-c)$$
A: 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, $$\frac{1}{2}(a^4+b^4+c^4) = (\frac{a^2 +b^2 +c^2}{2})^2$$
