## Inequality (sin, r and s)

For discussing Olympiad Level Algebra (and Inequality) problems
sowmitra
Posts: 155
Joined: Tue Mar 20, 2012 12:55 am
Location: Mirpur, Dhaka, Bangladesh

### Inequality (sin, r and s)

Prove the inequality,
$\frac{1}{\sqrt{2\sin A}}+\frac{1}{\sqrt{2\sin B}}+\frac{1}{\sqrt{2\sin C}}\leq \sqrt{\frac{s}{r}}$
where, $s$ and $r$ are the semi-perimeter and inradius of $\triangle ABC$.

Sharygin Geometry Olympiad, Russia
"Rhythm is mathematics of the sub-conscious."
Some-Angle Related Problems;

Nirjhor
Posts: 136
Joined: Thu Aug 29, 2013 11:21 pm
Location: Varies.

### Re: Inequality (sin, r and s)

The inequality simplifies to $\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}\le \dfrac{a+b+c}{\sqrt{abc}}\Rightarrow \sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le a+b+c$ after making the substitutions $2\sin A=\dfrac a R$ etc, $r=\dfrac{abc}{4sR}$, and $2s=a+b+c$.

This is a standard AM-GM result with equality at $a=b=c$ i.e. equilateral triangle.

A lower bound is $3^{3/4}$ found immediately via Jensen's, equality at equilateral triangle.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.

Tahmid
Posts: 110
Joined: Wed Mar 20, 2013 10:50 pm

### Re: Inequality (sin, r and s)

my solution is too much long

$\frac{1}{\sqrt{2sinA}}+\frac{1}{\sqrt{2sinB}}+\frac{1}{\sqrt{2sinC}}\leq \sqrt{\frac{s}{r}}$
or,$\sqrt{\frac{R}{a}}+\sqrt{\frac{R}{b}}+\sqrt{\frac{R}{c}}\leq \sqrt{\frac{s}{r}}$
or,$\frac{R}{a}+\frac{R}{b}+\frac{R}{c}+\frac{2R}{\sqrt{ab}}+\frac{2R}{\sqrt{bc}}+\frac{2R}{\sqrt{ca}}\leq \frac{s}{r}$
or,$R(\frac{ab+bc+ca}{abc})+2R(\frac{a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}}{abc})\leq \frac{s}{r}$
or,$\frac{R}{abc}(ab+bc+ca)+\frac{2R}{abc}(a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab})\leq \frac{s^{2}}{\Delta }$
or,$\frac{1}{4\Delta}(ab+bc+ca)+\frac{1}{2\Delta}(a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab})\leq \frac{(a+b+c)^{2}}{4\Delta }$
or,$(ab+bc+ca)+2(a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab})\leq (a+b+c)^{2}$
or,$ab+bc+ca+2a\sqrt{bc}+2b\sqrt{ca}+2c\sqrt{ab}\leq a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$
or,$2a\sqrt{bc}+2b\sqrt{ca}+2c\sqrt{ab}\leq a^{2}+b^{2}+c^{2}+ab+bc+ca$

now from AM-GM we know , $\frac{a^{2}+bc}{2}\geq \sqrt{a^{2}bc} \Leftrightarrow a^{2}+bc\geq 2a \sqrt{bc}$
samely , $b^{2}+ca\geq 2b \sqrt{ca}$ and $c^{2}+ab\geq 2c \sqrt{ab}$

by adding then we have $2a\sqrt{bc}+2b\sqrt{ca}+2c\sqrt{ab}\leq a^{2}+b^{2}+c^{2}+ab+bc+ca$

so, we are done