# Dorin Marghidanu's Inequality with Maximum Side

### Problem

In $\Delta ABC,\,$ $a=\max\{a,b,c\},\,$ $h_a\,$ is the altitude to side $a.\,$ Prove that $\displaystyle h_a \le \frac{p}{\sqrt{3}},\,$ where $p=\displaystyle \frac{a+b+c}{2}\,$ is the semiperimeter of the triangle.

### Solution 1

Let $\displaystyle \frac{A}{2}=x,\,$ $\displaystyle \frac{B}{2}=y,\,$ $\displaystyle \frac{C}{2}=z.\,$ Then $x,y,z\gt 0,\,$ $x+y+z=\displaystyle \frac{\pi}{2},\,$ and $x\ge y,\,$ $x\ge z,\,$ so that $x\ge\displaystyle \frac{\pi}{6}\,$ and $y+z\le\displaystyle \frac{\pi}{3}.$

It's known that $h_a=2R\sin 2y\sin 2z\,$ and $p=4R\cos x\cos y\cos z.\,$ We, thus need to prove that

$\displaystyle 2R\sin 2y\sin 2z\le\frac{4R\cos x\cos y\cos z}{\sqrt{3}}.$

Using the double argument formulas, this reduces to

$\displaystyle 2\sin y\sin z\le\frac{\cos x}{\sqrt{3}}.$

But $\displaystyle \sin y\sin z\le\sin^2\displaystyle \left(\frac{y+z}{2}\right).\,$ Thus, suffice it to prove that $\displaystyle 2\sin^2\left(\frac{y+z}{2}\right)\le\frac{\cos x}{\sqrt{3}}.\,$ We have a sequence of equivalences:

\displaystyle \begin{align} &2\sin^2\left(\frac{y+z}{2}\right)\le\frac{\cos x}{\sqrt{3}}\,\Longleftrightarrow\, 2\sin^2\left(\frac{y+z}{2}\right)\le\frac{\sin (y+z)}{\sqrt{3}}\,\Longleftrightarrow\\ &\qquad\sin\left(\frac{y+z}{2}\right)\le\frac{\displaystyle \cos \left(\frac{y+z}{2}\right)}{\sqrt{3}}\,\Longleftrightarrow\,\tan\left(\frac{y+z}{2}\right)\le\frac{1}{\sqrt{3}}. \end{align}

which is true because $\displaystyle \frac{x+y}{2}\le\frac{\pi}{6}.$

### Acknowledgment

The problem has been posted by Dorin Marghidanu at the mathematical inequalities facebook group where it gathered several responses. Subsequently, Prof Marghidanu posted the problem at the CutTheKnotMath facebook page with a whole collection of proofs and Prof Dinca has communicated his solution via the facebook messenger. Solution 1 is by Leo Giugiuc; Solution 2 is by Dorin Marghidanu; Solution 3 is by Vaggelis Stamatiadis; Solution 4 is by Diego Alvariz; Solution 5 is by Marian Dinca; Solution 6 is by Richdad Phuc.

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