Triangle area inequalities; semiperimeter

Question

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!

Answer 1

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:

$$ 12S\leq a^2+b^2+a^2+c^2+c^2+b^2\\ 12S\leq 2a^2+2b^2+2c^2\\ 6S\leq a^2+b^2+c^2 $$

Answer 2

Have you considered the well-known Heron's formula for computing the area $ S $ of a given triangle (see Heron's formula), namely $$ S=\frac{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{4}?$$

Answer 3

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 $K=\color{red}{\frac{1}{4\sqrt{3}}}.$