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.

About YiMin

This is just a nerd PhD student of Math@UT Austin.