Marian Dinca's Inequality

Problem

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 $C(B,BC).$

Marian Dinca's Inequality  - proof 2, 1

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 $E.$

Marian Dinca's Inequality  - proof 2, 2

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

Marian Dinca's Inequality  - proof 2, 3

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.

 

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