<s>[INST] Triangle area inequalities; semiperimeter I've got stuck on this problem :

Proof that for every triangle of
sides $a$, $b$ and $c$ and area
$S$, the  following
inequalities are true :
$4S \le a^2 + b^2$
$4S \le b^2 + c^2$
$4S \le a^2 + c^2$
$6S \le a^2 + b^2 + c^2$

The first thing that came to my mind was the inequality $S \le \frac 12 ab$. That is derived from the fact that $S = \frac 12ab \cdot \sin(\angle ACB)$ and $0 < \sin(\angle ACB) \le 1$.
Anyway, this wasn't enough to solve the problem. Some help would be well received.
Thanks!
 [/INST] Those inequalities are very weak. It is trivial that
$$ 2S\leq ab\leq \left(\frac{a+b}{2}\right)^2\leq \frac{a^2+b^2}{2}$$
by the GM-AM-QM inequality, hence $S\leq\frac{a^2+b^2+c^2}{6}$ by adding the first three inequalities.
However, they cannot all hold as equalities, since a (Euclidean) triangle cannot have three right angles, so it is reasonable to expect that $S\leq K(a^2+b^2+c^2)$ always holds with a constant $K<\frac{1}{6}$. By Heron's formula:
$$ 4S=\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)} $$
but:
$$ a^4+b^4+c^4 = 3\cdot QM(a^2,b^2,c^2)^2 \geq 3\cdot AM(a^2,b^2,c^2)^2 = \frac{1}{3}(a^2+b^2+c^2)^2 $$
hence the optimal constant (achieved by the equilateral triangle) is $K=\color{red}{\frac{1}{4\sqrt{3}}}.$
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