<s>[INST] 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.
 [/INST] 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.
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