Proving $\left(a^2+b^2\right)^2\geqslant(a+b+c)(a+b-c)(b+c-a)(c+a-b)$ for positive reals $a$, $b$, $c$ Question $5$ of BMO1 $2008$:
For positive real numbers $\;a,\;b,\;c,\;$ prove that $$\left(a^2+b^2\right)^2\geqslant(a+b+c)(a+b-c)(b+c-a)(c+a-b)$$
I noticed that the right side can be grouped, but did not get further.

Expanding both sides and moving positive terms to their respective sides, we get:
$a^4 +b^4+\dfrac{c^4}{2}\geqslant a^2 c^2 + b^2 c^2$
By AMGM we have:
$a^4 + \dfrac{c^4}{4}\geqslant a^2 c^2\;$ and $\;b^4 + \dfrac{c^4}{4}\geqslant b^2 c^2$
