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.

Let's simplify the inequality:
$$(a^2 + b^2)^2 ≥ (a + b + c)(a + b − c)(b + c − a)(c + a − b) \Rightarrow \\
a^4+2a^2b^2+b^4\ge ((a+b)^2-c^2)(c^2-(a-b)^2) \Rightarrow \\
a^4+2a^2b^2+b^4\ge 2c^2(a^2+b^2)-(a^2-b^2)^2-c^4 \Rightarrow \\
c^4-2(a^2+b^2)c^2+2(a^4+b^4)\ge 0 \qquad (1)$$
It is a bi-quadratic inequality and its discriminant is:
$$D=(a^2+b^2)^2-2(a^4+b^4)=2a^2b^2-(a^4+b^4)\le 0,$$
the iquality occurs for $a=b$. Note that the inequlity $(1)$ is true for $D<0$. We will check $D=0$. Then the inequality $(1)$ and its solution will be:
$$\begin{cases} c^2=a^2+b^2=2a^2 \\ 
(2a^2)^4-4a^4+4a^4\ge 0\end{cases} \Rightarrow16a^8\ge 0.$$
