Dorin Marghidanu's Inequality with Maximum Side

Source

Dorin Marghidanu's Inequality with Maximum Side, source

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 $\displaystyle \frac{x+y}{2}\le\frac{\pi}{6}.$

Solution 2

From $a=\max\{a,b,c\},\,$ 3a\ge a+b+c=2p,\,$ implying $\displaystyle 1\le\frac{3}{2}\cdot\frac{a}{p}.\,$ Using the means inequality,

$\displaystyle \begin{align} h_a &= 1\cdot h_a \le\frac{3}{2}\cdot\frac{a}{p}\cdot h_a\\ &\frac{3}{p}[\Delta ABC]=\frac{3}{p}\sqrt{p(p-a)(p-b)(p-c)}\\ &=\frac{3}{\sqrt{p}}\sqrt{(p-a)(p-b)(p-c)}\\ &=\frac{3}{\sqrt{p}}\sqrt{\left(\frac{(p-a)+(p-b)+(p-c)}{3}\right)^3}\\ &=\frac{3}{\sqrt{p}}\sqrt{\left(\frac{3p-2p}{3}\right)^3}\\ &=\frac{3}{\sqrt{p}}\sqrt{\left(\frac{p}{3}\right)^3}\\ &=\frac{p}{\sqrt{3}}. \end{align}$

Equality holds when $p-a=p-b=p-c,\,$ i.e., when $a=b=c.$

Solution 3

$\displaystyle \begin{align} (p-b)+(p-c)&\overset{AM-GM}{=}2\sqrt{(p-b)(p-c)}\,\Longrightarrow\, a\ge 2\sqrt{(p-b)(p-c)}\,\Longrightarrow\\ \sqrt{p(p-a)}&\ge \frac{2}{a}\sqrt{p(p-a)(p-b)(p-c)}\,\Longrightarrow\, \sqrt{p(p-a)}\ge h_a. \end{align}$

So, it is enough to prove that $\displaystyle \sqrt{p(p-a)}\le\frac{\sqrt{3}}{3}.\,$ We have a series of equivalences:

$\displaystyle \begin{align} \sqrt{p(p-a)}\le\frac{\sqrt{3}}{3}\,\Longleftrightarrow\, p(p-a)\le\frac{p^2}{3}\,\Longleftrightarrow\\ 3(p-a)\le p\,\Longleftrightarrow\,2p-3a\le 0\,\Longleftrightarrow\,3a-2p\ge 0\,\Longleftrightarrow\\ 3a-a-b-c\ge 0\,\Longleftrightarrow\,(a-b)+(a-c)\ge 0 \end{align}$

which is true because $a=\max\{a,b,c\}.$

Solution 4

We know that $a+b+c\ge 6\sqrt{3}r,\,$ where $r\,$ is the inradius of $\Delta ABC.\,$ Now $a+a+a\ge a+b+c\,$ so that $a\ge 2\sqrt{3}{r},\,$ implying $pa\ge 2\sqrt{3}pr=2\sqrt{3}\Delta,\,$ where $\Delta-[\Delta ABC],\,$ the area of $\Delta ABC.\,$ Thus. $\displaystyle p\ge 2\sqrt{3}\frac{\Delta}{a}=\sqrt{3}h_a.\,$ In other words, $\displaystyle h_a\le\frac{1}{\sqrt{3}}p.$

Solution 5

Since $a=\max\{a,b,c\},\,$ by Euclid I.18, $A=\max\{A,B,C\},\,$ so that $\displaystyle A\ge\frac{\pi}{3}.$ By the Law of Sines, with $R\,$ the circumradius, we have

$\displaystyle\begin{align}h_a&\le\frac{\sqrt{3}}{3}p\,\Longleftrightarrow\\ 2R\sin B\sin C&\le\frac{\sqrt{3}}{3}R(\sin A+\sin B+\sin C)\,\Longleftrightarrow\\ 2\sqrt{3}&\le\frac{\sin A}{\sin B\sin C}+\frac{1}{\sin C}+\frac{1}{\sin B}. \end{align}$

Now, by Bergstrom's inequality,

$\displaystyle \frac{1}{\sin C}+\frac{1}{\sin B}\ge\frac{(1+1)^2}{\sin A+\sin B}\ge\frac{4}{\displaystyle 2\sin\left(\frac{C+B}{2}\right)}=\frac{2}{\displaystyle \cos\frac{A}{2}}.$

On the other hand, $\displaystyle \sin B\sin C\le\sin^2\left(\frac{B+C}{2}\right)=\cos^2\left(\frac{A}{2}\right)\,$ such that, finally,

$\displaystyle \begin{align} \frac{\sin A}{\sin B\sin C}+\frac{1}{\sin C}+\frac{1}{\sin B}&\ge\frac{\sin A}{\displaystyle \cos\left(\frac{A}{2}\right)}+\frac{2}{\displaystyle \cos\frac{A}{2}}\\ &=\frac{\displaystyle 2\sin\frac{A}{2}}{\displaystyle \cos\frac{A}{2}}+\frac{2}{\displaystyle \cos\frac{A}{2}}\\ &=\frac{\displaystyle 2\sin\frac{A}{2}+2}{\displaystyle \cos\frac{A}{2}}. \end{align}$

Function $\displaystyle f(A)=\frac{\displaystyle 2\sin\frac{A}{2}+2}{\displaystyle \cos\frac{A}{2}}\,$ is increasing for $\displaystyle A\in\left(\frac{\pi}{3},\pi\right)\,$ because $\displaystyle \sin\frac{A}{2}\,$ is increasing while $\displaystyle \cos\frac{A}{2}\,$ is decreasing in that interval. It follows that

$\displaystyle f(A)\ge f\left(\frac{\pi}{3}\right)=\frac{\displaystyle 2\cdot\frac{1}{2}+2}{\displaystyle \frac{\sqrt{3}}{2}}=2\sqrt{3}.$

The latter implies the required inequality.

Solution 6

We shall prove more, viz., $\displaystyle m_a\le\frac{p}{\sqrt{3}},\,$ where $m_a\,$ is the median to $a.\,$ We have

$\displaystyle \begin{align} m_a&\le\frac{p}{\sqrt{3}}\,\Longleftrightarrow\\ \frac{2(b^2+c^2)-a^2}{4}&\le\frac{(a+b+c)^2}{12}\,\Longleftrightarrow\\ 0&\le 4a^2+2a(b+c)+2bc-5(b^2+c^2)\overset{def}{=}f(a). \end{align}$

Since $a=\max\{a,b,c\},\,$ let $t=2a-(b+c)\gt 0,\,$ and $2a=t+b+c.\,$ Thus,

$\displaystyle \begin{align} t&\ge|b-c|\,\Longleftrightarrow\\ [2a-(b+c)]^2&\ge (b-c)^2\,\Longleftrightarrow\\ (2a-2b)(2a-2c)^\ge 0, \end{align}$

which is true. Now,

$f(a)=(t+b+c)^2+(t+b+c)(b+c)+2bc-5(b^2+c^2),$

and $f(a)=t^2+3t(b+c)-3(b-c)^2\ge 0,\,$ because $t(b+c)\ge (b-c)^2.\,$ Equality is attained when $a=b=c.\,$ Now, since $m_a\ge h_a\,$ we also proved the required inequality.

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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