Misael Matthews

2022-06-25

Prove this inequality where a, b and c are sides of a triangle and S its Area.

$\frac{ab+bc+ca}{4S}\ge \mathrm{ctg}\frac{\pi}{6}$

$\frac{ab+bc+ca}{4S}\ge \mathrm{ctg}\frac{\pi}{6}$

Nola Rivera

Beginner2022-06-26Added 21 answers

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 $\frac{1}{\mathrm{sin}(x)}$ is a convex function on the interval $(0,\pi )$, (1) is a straightforward consequence of Jensen's inequality.

$\begin{array}{}\text{(1)}& \frac{1}{\mathrm{sin}A}+\frac{1}{\mathrm{sin}B}+\frac{1}{\mathrm{sin}C}\ge 2\sqrt{3}\end{array}$

and since $\frac{1}{\mathrm{sin}(x)}$ is a convex function on the interval $(0,\pi )$, (1) is a straightforward consequence of Jensen's inequality.

Damon Stokes

Beginner2022-06-27Added 6 answers

Since $ab+ac+bc\ge \sum _{cyc}(2ab-{a}^{2})$ it's just $\sum _{cyc}(a-b{)}^{2}\ge 0$ and

$\sum _{cyc}(2ab-{a}^{2})=\sum _{cyc}a(b+c-a)>0,$

it's enough to prove that

$\sum _{cyc}(2ab-{a}^{2})\ge 4\sqrt{3}S$

or

$\sum _{cyc}(2ab-{a}^{2})\ge \sqrt{3\sum _{cyc}(2{a}^{2}{b}^{2}-{a}^{4})}$

or

$\sum _{cyc}({a}^{4}-{a}^{3}b-{a}^{3}c+{a}^{2}bc)\ge 0,$

which is Schur.

Done!

$\sum _{cyc}(2ab-{a}^{2})=\sum _{cyc}a(b+c-a)>0,$

it's enough to prove that

$\sum _{cyc}(2ab-{a}^{2})\ge 4\sqrt{3}S$

or

$\sum _{cyc}(2ab-{a}^{2})\ge \sqrt{3\sum _{cyc}(2{a}^{2}{b}^{2}-{a}^{4})}$

or

$\sum _{cyc}({a}^{4}-{a}^{3}b-{a}^{3}c+{a}^{2}bc)\ge 0,$

which is Schur.

Done!

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