# Leo Giugiuc's Inequality in Triangle, Solely with Cotangents

### Solution 1

We start with

Lemma

If $A\in (0,\displaystyle \frac{\pi}{4}],\,$ then

(1)

$(2\sqrt{2}-1)\sin A+\cos A\le 2.$

Indeed, from $A\in (0,\displaystyle \frac{\pi}{4}],\,$ $\cos A\ge \displaystyle \frac{\sqrt{2}}{2}\,$ while (1) is equivalent to

\displaystyle \begin{align} &2-\cos A\ge (2\sqrt{2}-1)\sin A\\ \Leftrightarrow\,&\,(4-4\cos A+\cos^2A\ge(9-4\sqrt{2})(1-\cos^2A)\\ \Leftrightarrow\,&\,(10-4\sqrt{2})\cos^2A-4\cos A-5+4\sqrt{2}\ge 0\\ \Leftrightarrow\,&\,(10-4\sqrt{2})\left(\cos A-\frac{\sqrt{2}}{2}\right)\left(\cos A-\frac{8-5\sqrt{2}}{10-4\sqrt{2}}\right)\ge 0, \end{align}

which is true because $\displaystyle \cos A\ge\frac{\sqrt{2}}{2}\gt \frac{8-5\sqrt{2}}{10-4\sqrt{2}}.$

We have $\displaystyle \cot A+\cot B+\cot C=\frac{a^2+b^2+c^2}{4S},\,$ where $S=[\Delta ABC],\,$ the area of $\Delta ABC,\,$ $4S=2bc\sin A\,$ and $a^2=b^2+c^2-2bc\cos A,\,$ combining which

$\displaystyle \cot A+\cot B+\cot C=\frac{b^2+c^2-bc\cos A}{bc\sin A}.$

It follows that $\cot A+\cot B+\cot C\ge 2\sqrt{2}-1\,$ is equivalent to

$\displaystyle \frac{b^2+c^2-bc\cos A}{bc\sin A}\ge 2\sqrt{2}-1,$

or,

(2)

$\displaystyle \frac{b^2+c^2}{bc}\ge(2\sqrt{2}-1)\sin A+\cos A.$

But we have

(3)

$\displaystyle \frac{b^2+c^2}{bc}\ge 2.$

So, (1) & (3) imply (2), completing the proof.

Equality holds iff $\displaystyle A=\frac{\pi}{4}\,$ and $\displaystyle B=C=\frac{3\pi}{8}.$

### Solution 2

The problem reduces to

If $x,y,z\gt 0\,$ and $xy+yz+zx=1,\,$ then $x+y+z\ge 2\sqrt{2}-1.$

We have $\displaystyle 1\le xx=\frac{1-yz}{y+z},\,$ implying $1-yz\ge y+z.\,$ But, by the AM-GM inequality, $y+z\ge2\sqrt{yz},\,$ so that $1-yz\ge 2\sqrt{yz},\,$ or $yz\le (\sqrt{2}-1)^2,\,$ with equality iff $y=z=\sqrt{2}-1.$

However,

$\displaystyle y+z=\frac{1-yz}{x}\ge\frac{1-(\sqrt{2}-1)^2}{x}=\frac{2\sqrt{2}-2}{x}.$

Hence, suffice it to prove that $\displaystyle x+\frac{2\sqrt{2}-2}{x}\ge 2\sqrt{2},\,$ or, equivalently, $(x-1)[x-(2\sqrt{2}-2)]\ge 0,\,$ which is true since $x\ge 1.$

Note that from the above proof we deduce that the equality holds for the triangle with $A=45^{\circ}\,$ and $B=C=67.5^{\circ}.$

### Acknowledgment

Leo Giugiuc has kindly posted the problem and his solution (Solution 2) at the CutTheKnotMath facebook page. Gabi Cuc Cucoanes commented with his own solution (Solution 1) and then also messaged me his solution. I am grateful to both of them.

Copyright © 1996-2017 Alexander Bogomolny

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