An Inequality in Triangle with Side Lengths and Circumradius

Problem

An Inquality in Triangle with Side Lengths and Circumradius  - problem

Solution 1

By the AM-GM inequality,

$\displaystyle\sum_{cycl}\frac{ab}{\sqrt{a^2+b^2}}\le\sum_{cycl}\frac{ab}{\sqrt{2ab}}=\frac{1}{\sqrt{2}}\sum_{cycl}\sqrt{ab}.$

By the Cauchy-Schwarz inequality,

$\displaystyle\frac{1}{\sqrt{2}}\sum_{cycl}\sqrt{ab}\le\frac{1}{\sqrt{2}}\sqrt{\sum_{cycl}a\cdot\sum_{cycl}b}=\frac{1}{\sqrt{2}}\cdot 2s,$

where $s\,$ is the semiperimeter of the triangle. By Mitrinović's inequality,

$\displaystyle\frac{1}{\sqrt{2}}\cdot 2s\le\sqrt{2}\cdot\frac{3\sqrt{3}}{2}R=\frac{3\sqrt{6}R}{2}.$

Solution 2

$\displaystyle\begin{align} &s=R(\sin A+\sin B+\sin C)\le\frac{3\sqrt{3}}{2}R,\\ &\frac{3\sqrt{3}}{2}R\ge s =\frac{2(a+b+c)}{4}\\ &\qquad\qquad=\sum_{cycl}\frac{a+b}{4}=\sum_{cycl}\frac{(a+b)^2}{4(a+b)}\\ &\qquad\qquad=\sum_{cycl}\left(\frac{a+b}{2}\right)^2\cdot\frac{1}{a+b}\ge \sum_{cycl}\frac{ab}{a+b}\\ &\qquad\qquad\ge\sum_{cycl}\frac{ab}{\sqrt{2(a^2+b^2)}}. \end{align}$

The required inequality follows.

Solution 3

From the AM-GM inequality, $\displaystyle\frac{ab}{\sqrt{a^2+b^2}}\le\frac{\sqrt{ab}}{\sqrt{2}}.$ Thus

$\displaystyle\sum_{cycl}\frac{ab}{\sqrt{a^2+b^2}}\le\frac{1}{\sqrt{2}}\sum_{cycl}\sqrt{ab}\le\frac{1}{\sqrt{2}}\sum{cycl}a.$

Further,

$\displaystyle\begin{align} &\frac{\sqrt{2}}{2}\sum_{cycl}a\le\frac{3\sqrt{6}R}{2}\,&\Leftrightarrow\\ &2R(\sin +\sin B+\sin C)\le 3\sqrt{3}R\,&\Leftrightarrow\\ &\frac{\sin A+\sin B+\sin C}{3}\le\frac{\sqrt{3}}{2}, \end{align}$

which is true because function $\sin:\,[0,\pi]\rightarrow\mathbb{R}\,$ is concave, implying, by Jensen's inequality,

$\displaystyle\frac{\sin A+\sin B+\sin C}{3}\le\sin\left(\frac{A+B+C}{3}\right)=\frac{\sqrt{3}}{2}.$

Solution 4

We assume a known inequality in a triangle $a^2+b^2+c^2\leq 9R^2.$

$\displaystyle\begin{align} LHS &= \frac{ab}{\sqrt{a^2+b^2}}+\frac{bc}{\sqrt{b^2+c^2}}+\frac{ca}{\sqrt{c^2+a^2}} \\ &\leq\frac{1}{2}(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})~\text{(AM-GM)} \\ &\leq \frac{\sqrt{6}}{2}\sqrt{a^2+b^2+c^2}~\text{(Jensen's)} \\ &\leq \frac{3\sqrt{6}}{2}R~\text{(assumed inequality)}. \end{align}$

Acknowledgment

Dan Sitaru has kindly posted the problem, due to Boris Colakovic, along with three solution3 at the CutTheKnotMath facebook page. Solution 1 is by Dan Sitaru himself; Solution 2 is by Yadamsuren Myagmarsuren; Solution 3 is by Geanina Tudose, Solution is by Amit Itagi.

 

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