# A Geometry Inequality

I checked my mailbox and found a post which was buried long time ago.

It was about a geometry inequality, and it is really interesting.

$\displaystyle\sum{\frac{1}{A}}\ge\frac{9}{2\pi}\sqrt{\frac{R}{2r}}$

where the symbol $\sum$ represents cyclic sum.

The proof was elementary now. But I still like this, full of magic skills.

Proof:

First we introduce two formulae of triangular geometry.

$2R^2\sin A\sin B\sin C=S_{\Delta}$,

$rp=S_{\Delta}$,

where the symbols’ meanings are obvious, I suppose.

Then the inequality turns into:

$\displaystyle{\sum\frac{1}{A}}\ge\frac{9}{2\pi}\sqrt{2\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}}$,

however, we may prove a simpler result by using:

$\displaystyle{\sum\frac{1}{A}}\ge\sqrt[3]{\Pi\frac{1}{A}}$.

Then we have to prove the function:

$H(A,B,C)=\displaystyle \frac{1}{(ABC)^2}-\frac{3^6}{8\pi^6}(\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}})^3\ge 0$

Fortunately, we can use Jensen Inequality here, we want to prove:

$1\displaystyle\ge\alpha\Pi A^2 (\sin{\frac{A}{2}})^3$,

which means we need find the maximum of the RHS.

We set, $\displaystyle f(A)=2\log A+3\log\sin{\frac{A}{2}}$, then,

$\displaystyle f''(A)=-\frac{2}{A^2}-\frac{3}{4}\frac{1}{\sin^2{\frac{A}{2}}}\textless 0$

Thus, $f$ is a convex function.

$\displaystyle\sum f(A)\le 3f(\frac{A+B+C}{3})=3 f(\frac{\pi}{3})$.

It is proven.

$\Box$

It is a very strong inequality actually.