# Marian Dinca's Inequality

### Problem

### Solution 1

Let $a+b+c=6$ and $a^2+b^2+c^2=6(2+t^2),\,$ $t\ge 0.\,$ Since $a,b,c$ are the side lengths of a triangle, $t\lt 1.\,$ Now, $4m_a^2=2(b^2+c^2)-a^2=12(2+t^2)-3a^2,\,$ implying that $\displaystyle m_a=\sqrt{3(2+t^2)-\frac{3a^2}{4}}.$ Thus, $m_a\le\sqrt{3}$ is equivalent to $a\ge 2\sqrt{1+t^2}.\,$

$a=\max\{a,b,c\}$ implies $2+t\le a.\,$ Suffice it to show that $2+t\lt 2\sqrt{1+t^2}\,$ which is equivalent to $t(4-3t)\ge 0,\,$ which is true since $0\le t\le 1.$

### Solution 2

Since $a$ is assumed the longest side of $\Delta ABC,\,$ $A$ lies on the ellipse with foci at $B$ and $C.$ In addition, $A$ belongs to the arc $DE$ of the ellipse (in its reflection), where $D$ is the intersection of the ellipse with the circle $C(C,BC)\,$ with center $C$ and radius $BC.\,$ Similarly, $E\,$ is the intersection of the ellipse with $C(B,BC).$

Under these conditions, and for the given $a$ and the semiperimeter $s,$ the median $m_a=AM$ attains its maximum when $A$ coincides with either $D$ or $E.$

It follows that we may restrict our attention to the case where $\Delta ABC$ is isosceles with $AC=BC=a.$

Let $\beta=\angle ABC=\angle BAC.\,$ By Euclid I.18, $60^{\circ}\le\beta\le 90^{\circ}.$

By the Law of Cosines in $\Delta ABC,\,$ $AC^2=BC^2+AB^2-2BC\cdot AB\cdot\cos\beta,\,$ from which $c=AB=2a\cos\beta.$ Thus $s=\displaystyle \frac{2a+c}{2}=a(1+\cos\beta).$ On the other hand,

$\displaystyle m_a^2=\frac{b^2+a^2}{2}-\frac{a^2}{4}=\frac{a^2}{4}(1+8\cos^2\beta).$

We thus need to establish that $\displaystyle \frac{m_a^2}{s^2}=\frac{1+8\cos^2\beta}{4(1+\cos\beta)^2}\le\frac{1}{3}.$

Setting, $u=\cos\beta,\,$ suffice it to verify that $20u^2-8u-1\le 0.\,$ The quadratic form on the left has the roots $\displaystyle u_{1,2}=\frac{4\pm 6}{20},\,$ so it's not positive for $\displaystyle u\in\left[-\frac{1}{10},\frac{1}{2}\right]$ which is true because, with $\beta\in [60^{\circ},90^{\circ}],\,$ $\displaystyle \cos\beta\in\left[0,\frac{1}{2}\right].$

Clearly, for $\beta=60^{\circ},\,$ $\displaystyle \frac{m_a^2}{s^2}=\frac{1}{3}.$

### Solution 3

The inequality is equivalent to

$(a+b+c)^2-3(2b^2+2c^2-a^2)\geq 0.$

$a=max\{a,b,c\}$. WLOG, let us assume $a\geq b \geq c$.

$\begin{align} &(a+b+c)^2-3(2b^2+2c^2-a^2) \\ &=4a^2-5b^2-5c^2+2(ab+bc+ca) \\ &\geq 4a^2-5b^2-5c^2+2(b^2+2ac)~\text{(Rearrangement)} \\ &\geq 4a^2-5b^2-5c^2+2(b^2+2c^2) \\ &\geq3(a^2-b^2)+(a^2-c^2)\geq 0. \end{align}$

### Solution 4

Set $a=x+y,$ $b=y+z,$ $c=x+z,$ $x\ge y\ge z\ge 0,$ $\displaystyle 2m_a^2=(x+z)^2+(y+z)^2-\frac{(x+y)^2}{2},\,$ $s=x+y+z.$

The required inequality translates into $3m_a^2\le s^2$ which is

$8z^2+4xz+4yz\le x^2+y^2+14xy,$

or, termwise,

$(z^2-x^2)+(z^2-y^2)+6(z^2-6xy)+4x(z-y)+4yz(z-x)\le 0$

since, by our assumption, none of the terms is positive.

### Acknowledgment

The problem, that is due to Marian Dinca, has been posted at the CutTheKnotMath facebook page by Leo Giugiuc, along with his solution (Solution 1); Solution 3 is by Amit Itagi, Solution 4 by Lorenzo Villa.

[an error occurred while processing this directive]

|Contact| |Up| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny