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!

Let $S$ denote the area of the triangle, so:
$$
S = \frac{ab\sin{(\gamma)}}{2} = \frac{cb\sin{(\alpha)}}{2} = \frac{ac\sin{(\beta)}}{2}
$$
And as for $\theta \in[0,\pi]$: $0\leq\sin{\theta}\leq1$, we get as you said:
$$
S\leq\frac{ab}{2}\\
S\leq\frac{ac}{2}\\
S\leq\frac{bc}{2}\\
$$
So:
$$
4S \leq 2ab\\
4S\leq 2ac\\
4S\leq 2bc\\
$$
And for $a,b,c\in {\mathbb{R}}^{+}$:
$$
{(a-b)}^{2}\geq 0\\
{(a-c)}^{2}\geq 0\\
{(c-b)}^{2}\geq 0
$$
We get:
$$
a^2+b^2\geq 2ab\\
a^2+c^2\geq 2ac\\
c^2+b^2\geq 2bc
$$
So we get:
$$
4S \leq a^2+b^2\\
4S\leq a^2+c^2\\
4S\leq c^2+b^2\\
$$
For the last one add the latter inequalities just obtained together:
$$
12S\leq a^2+b^2+a^2+c^2+c^2+b^2\\
12S\leq 2a^2+2b^2+2c^2\\
6S\leq a^2+b^2+c^2
$$
