British Maths Olympiad (BMO) 2006 Round 1 Question 5, alternate solution possible? The question states 
For positive real numbers $a,b,c$ prove that
$(a^2 + b^2)^2 ≥ (a + b + c)(a + b − c)(b + c − a)(c + a − b)$
After some algebraic wrangling we can get to the point where:
$(a^2 + b^2)^2 + (a + b)^2(a − b)^2 + c^4 ≥  2c^2(a^2 + b^2)$
At this point if we take the $LHS - RHS$ we can write the expression as the sum of squares proving the inequality.
I was wondering, is it possible to divide both sides by $c^2(a^2 + b^2)$ and show somehow that 
$((a^2 + b^2)^2 + (a + b)^2(a − b)^2 + c^4)/(c^2(a^2 + b^2)) ≥  2$ 
I tried but was not able to.

We need to prove that
$$(a^2+b^2)^2\geq\sum_{cyc}(2a^2b^2-a^4)$$ or
$$c^4-2(a^2+b^2)c^2+2(a^4+b^4)\geq0$$ or
$$(c^2-a^2-b^2)^2+(a^2-b^2)^2\geq0.$$
Yes, you can  prove this inequality by the dividing. 
Indeed, if $c^2(a^2+b^2)=0$ then the inequality is obvious.
Let $c^2(a^2+b^2)\neq0$.
Thus, by AM-GM and Cauchy-Schwarz
$$\frac{c^4+2(a^4+b^4)}{c^2(a^2+b^2)}=\frac{c^2}{a^2+b^2}+\frac{2(a^4+b^4)}{c^2(a^2+b^2}\geq$$
$$\geq2\sqrt{\frac{c^2}{a^2+b^2}\cdot\frac{2(a^4+b^4)}{c^2(a^2+b^2)}}=2\sqrt{\frac{2(a^4+b^4)}{(a^2+b^2)^2}}=$$
$$=2\sqrt{\frac{(1+1)(a^4+b^4)}{(a^2+b^2)^2}}\geq2\sqrt{\frac{(a^2+b^2)^2}{(a^2+b^2)^2}}=2.$$
