<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] 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.
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