Misael Matthews

2022-06-25

Prove this inequality where a, b and c are sides of a triangle and S its Area.
$\frac{ab+bc+ca}{4S}\ge \mathrm{ctg}\frac{\pi }{6}$

Nola Rivera

By the sine theorem, the given inequality is equivalent to
$\begin{array}{}\text{(1)}& \frac{1}{\mathrm{sin}A}+\frac{1}{\mathrm{sin}B}+\frac{1}{\mathrm{sin}C}\ge 2\sqrt{3}\end{array}$
and since $\frac{1}{\mathrm{sin}\left(x\right)}$ is a convex function on the interval $\left(0,\pi \right)$, (1) is a straightforward consequence of Jensen's inequality.

Damon Stokes

Since $ab+ac+bc\ge \sum _{cyc}\left(2ab-{a}^{2}\right)$ it's just $\sum _{cyc}\left(a-b{\right)}^{2}\ge 0$ and
$\sum _{cyc}\left(2ab-{a}^{2}\right)=\sum _{cyc}a\left(b+c-a\right)>0,$
it's enough to prove that
$\sum _{cyc}\left(2ab-{a}^{2}\right)\ge 4\sqrt{3}S$
or
$\sum _{cyc}\left(2ab-{a}^{2}\right)\ge \sqrt{3\sum _{cyc}\left(2{a}^{2}{b}^{2}-{a}^{4}\right)}$
or
$\sum _{cyc}\left({a}^{4}-{a}^{3}b-{a}^{3}c+{a}^{2}bc\right)\ge 0,$
which is Schur.
Done!

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