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.

Suppose that $a,b,c$ can form the sides of a triangle.  Let $s=\frac{a+b+c}{2}$ be the semiperimeter.  The inequality becomes $$
(a^2+b^2)^2\ge2s\cdot 2(s-a)\cdot 2(s-b)\cdot 2(s-c)
$$
or by Heron's formula, $$a^2+b^2\ge 4A$$
where $A$ is the area of the triangle.  If $\theta$ is the angle between sides $a$ and $b$, this reduces to $$
a^2-2ab\sin\theta+b^2\ge0,$$ and we have$$
a^2-2ab\sin\theta+b^2\ge a^2-2ab+b^2=(a-b)^2\ge0$$
In the case where $a,b,c$ do not form a triangle, exactly one of the factors on the right-hand is negative or one of the factors is $0$, so the inequality is trivial. 
