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.
 A: 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 $\;b^4 + \dfrac{c^4}{4}\geqslant b^2 c^2$
A: If the right hand side is negative then the inequality trivially holds. Otherwise $a,b,c$ are the sides of a triangle, and by Heron's formula for the area $S$ of a triangle the inequality reduces to:
$$
a^2+b^2 \ge 4 S
$$
This follows from $S = \frac{1}{2} ab \sin C \le \frac{1}{2} ab$ and the means inequalities $ab \le \left(\frac{a+b}{2}\right)^2\le \frac{a^2+b^2}{2}\,$.

[ EDIT ] It is not possible for two factors on the right hand side of the original expression to be negative, since for example $a+b-c\lt 0$ and $b+c-a \lt 0$ would imply $b \lt 0$. Therefore, when the right hand side is non-negative each factor must be non-negative, so $a,b,c$ observe the triangle inequalities and must be the sides of a (possibly degenerate) triangle.
A: Note that
$$
(a+b+c)(a+b-c)(a-b+c)(-a+b+c)=((a+b)^2-c^2)(c^2-(a-b)^2)=
\\
=(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)=(2ab)^2-(a^2+b^2-c^2)^2.
$$
Can you continue now?
A: And now, a geometric argument.
If any of the terms on the right hand side are nonpositive, the inequality follows trivially. If they are all positive, $a,b,c$ form the legs of a triangle:

By Heron's formula, the area of this triangle is
$$
\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}
$$
The area of the triangle is also half the sum of the area of the two rectangles. However, a rectangle with diagonal $d$ can't have more area than $d^2/2$, so the total area of the two rectangles is less than $(a^2 + b^2)/2$. Since the triangle has half this area, we have
$$
\frac{a^2 + b^2}{4} \ge \frac{1}{4}\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}
$$
Squaring both sides and clearing denominators then gives your inequality.
