# An Inequality in Triangle, VIII

### Problem

where $a,b,c\;$ are the side lengths of the triangle and $m_a,m_b,m_c\;$ are the lengths of its medians, as customary.

### Proof

Denote $m_a=x,\;$ $m_b=y,\;$ $m_c=z.\;$ We know that there is $\Delta XYZ\;$ with $YZ=x,\;$ $ZX=y,\;$ and $XY=Z.\;$ Further, from

\displaystyle\begin{align} 2(b^2+c^2)-a^2&=4x^2,\\ 2(c^2+a^2)-b^2&=4y^2,\\ 2(a^2+b^2)-c^2&=4z^2 \end{align}

we derive

\displaystyle\begin{align} a^2&=\frac{4}{9}[2(y^2+z^2)-x^2],\\ b^2&=\frac{4}{9}[2(z^2+x^2)-y^2],\\ c^2&=\frac{4}{9}[2(x^2+y^2)-z^2],\\ \end{align}

Thus, the required inequality reduces to

$\displaystyle\sum_{cycl}\left(\frac{y^2+z^2-x^2}{\sqrt{xy}}\right)\le x+y+z.$

This is equivalent to showing that

$\displaystyle 2\sqrt{yz}\cos X+2\sqrt{zx}\cos Y+2\sqrt{xy}\cos Z\le x+y+z,$

or,

$\displaystyle x-2(\sqrt{z}\cos Y+\sqrt{y}\cos Z)\sqrt{x}+(y-2\sqrt{yz}\cos X+z)\ge 0.$

Consider the quadratic polynomial $f(t)=t^2-2(\sqrt{z}\cos Y+\sqrt{y}\cos Z)t+(y-2\sqrt{yz}\cos X+z)\;$ on interval $(0,\infty).$

The discriminant:

\begin{align} \Delta&=(\sqrt{z}\cos Y+\sqrt{y}\cos Z)^2-(y-2\sqrt{yz}\cos X+z)\\ &=-y\sin^2Z-z\sin^2Y+2\sqrt{yz}(\cos Y\cos Z+\cos X)\\ &=-y\sin^2Z-z\sin^2Y+2\sqrt{yz}\sin Y\sin Z\\ &=-(\sqrt{y}\sin Z-\sqrt{z}\sin Y)^2\\ &\le 0, \end{align}

implying that $f(t)\ge 0,\;$ in particular, $f(\sqrt{x})\ge 0,\;$

i.e.,

$\displaystyle x-2(\sqrt{z}\cos Y+\sqrt{y}\cos Z)\sqrt{x}+(y-2\sqrt{yz}\cos X+z)\ge 0.$

### Acknowledgment

The inequality, which is due to Nguyen Hung Viet, has been kindly communicated to me by Leo Giugiuc, along with his proof.