# An Inequality in Triangle, with Sides and Medians III

### Solution 1

\displaystyle \begin{align} \sum_{cycl}(m_b+m_c-m_a)^2&=3\sum_{cycl}m_a^2-2\sum_{cycl}m_am_b\\ &=\frac{9}{4}\sum_{cycl}a^2-2\sum_{cycl}m_am_b\\ &\ge\frac{9}{4}\sum_{cycl}a^2-\frac{1}{2}\sum_{cycl}(2a^2+bc)\\ &\ge \frac{9}{4}\sum_{cycl}a^2-\frac{3}{2}\sum_{cycl}a^2=\frac{3}{4}\sum_{cycl}a^2. \end{align}

Again,

\displaystyle \begin{align} \sum_{cycl}m_a(m_b+m_c-m_a)&=2\sum_{cycl}m_am_b-\sum_{cycl}m_a^2\\ &\le\frac{1}{2}\sum_{cycl}(2a^2+bc)-\frac{3}{4}\sum_{cycl}a^2\\ &\le\frac{3}{2}\sum_{cycl}a^2=\frac{3}{4}\sum_{cycl}a^2. \end{align}

It follows, by Bergstrom's inequality, that

\displaystyle \begin{align} \sum_{cycl}\frac{(m_b+m_c-m_a)^3}{m_a}&=\sum_{cycl}\frac{(m_b+m_c-m_a)^4}{m_a(m_b+m_c-m_a)}\\ &\ge\frac{\displaystyle \left(\sum_{cycl}(m_b+m_c-m_a)^2\right)^2}{\displaystyle \sum_{cycl}m_a(m_b+m_c-m_a)}\\ &\ge\frac{\displaystyle \left(\frac{3}{4}(a^2+b^2+c^2)\right)^2}{\displaystyle \frac{3}{4}(a^2+b^2+c^2)}\\ &=\frac{3}{4}(a^2+b^2+c^2). \end{align}

### Solution 2

Let $m_b+m_c-m_a=x,\,$ $m_c+m_a-m_b=y,\,$ $m_a+m_b-m_c=z.\,$ Note that, $x,y,z\gt 0,\,$ $m_a+m_b+m_c=x+y+z,\,$ $\displaystyle m_a=\frac{y+z}{2},\,$ $\displaystyle m_b=\frac{z+x}{2},\,$ $\displaystyle m_c=\frac{x+y}{2}.\,$ We thus have:

$\displaystyle LHS=\frac{2x^3}{y+z}+\frac{2y^3}{z+x}+\frac{2z^3}{x+y},$

while

\displaystyle\begin{align}RHS&=\frac{3}{4}\sum_{cycl}a^2=\sum_{cycl}m_a^2\\ &=\frac{1}{4}\sum_{cycl}(y+z)^2=\frac{1}{2}\left(\sum_{cycl}x^2+\sum_{cycl}xy\right) \end{align}

such that the required inequality reduces to

$\displaystyle \sum_{cycl}\frac{x^3}{y+z}\ge \frac{1}{4}\left(\sum_{cycl}x^2+\sum_{cycl}xy\right).$

Note that $\displaystyle LHS=\sum_{cycl}\frac{x}{y+z}\cdot x^2.\,$ WLOG, $x\ge y\ge z,\,$ then $x+y\ge x+z\ge y+z,\,$ implying $\displaystyle \frac{x}{y+z}\ge\frac{y}{z+x}\ge\frac{z}{x+y}.\,$ Now we are in a position to apply Chebyshev's inequality:

\displaystyle\begin{align} LHS&\ge\frac{1}{3}\sum_{cycl}\frac{x}{y+z}\cdot\sum_{cycl}x^2\\ &=\frac{1}{3}\sum_{cycl}\frac{x^2}{xy+zx}\cdot\sum_{cycl}x^2\\ &\ge\frac{1}{3}\frac{\displaystyle \left(\sum_{cycl}x\right)^2}{\displaystyle 2\sum_{cycl}{xy}}\cdot\sum_{cycl}x^2. \end{align}

Thus, suffice it to prove that

$\displaystyle \frac{\displaystyle \left(\sum_{cycl}x^2+2\sum_{cycl}xy\right)\left(\sum_{cycl}x^2\right)}{\displaystyle 6\sum_{cycl}{xy}}\ge \frac{1}{4}\left(\sum_{cycl}x^2+\sum_{cycl}xy\right).$

With $\displaystyle u=\sum_{cycl}x^2\,$ and $\displaystyle v=\sum_{cycl}xy,\,$ the inequality rewrites as

$\displaystyle \frac{(u+2v)u}{3v}\ge \frac{u+v}{2}.$

This is successively equivalent to

\displaystyle\begin{align} \Leftrightarrow\,&\,2u^2+4uv\ge 3uv+3v^2\\ \Leftrightarrow\,&\,2u^2+uv-3v^2\ge 0\\ \Leftrightarrow\,&\,(u-v)(2u+3v)\ge 0 \end{align}

because, as we know, $\displaystyle u=\sum_{cycl}x^2\ge\sum_{cycl}xy=v.$

### Solution 3

If we take into account the well known expressions for the medians, we shall reduce the inequality to

$\displaystyle \sum_{cycl}\frac{(m_b+m_c-m_a)^3}{m_a}\ge m_a^2+m_b^2+m_c^2.$

For simplicity, let's use $x,y,z\,$ for $m_a,m_b,m_c.\,$ Thus the inequality to prove becomes

$\displaystyle \sum_{cycl}\frac{(x+y-z)^3}{z}\ge x^2+y^2+z^2.$

From here we follow in the footsteps of Solution 1. On one hand,

$\displaystyle \sum_{cycl}(x+y-z)^2=3\sum_{cycl}x^2-2\sum_{cycl}xy\ge\sum_{cycl}x^2.$

On the other hand,

$\displaystyle \sum_{cycl}z(x+y-z)=2\sum_{cycl}xy-\sum_{cycl}x^2\le\sum_{cycl}x^2.$

It follows by Bergstrom's inequality that

\displaystyle \begin{align} \sum_{cycl}\frac{(x+y-z)^3}{z}&=\sum_{cycl}\frac{(x+y-z)^4}{z(x+y-z)}\\ &\ge\frac{\displaystyle \left(\sum_{cycl}(x+y-z)^2\right)^2}{\displaystyle \sum_{cycl}z(x+y-z)}\\ &\ge\frac{\displaystyle \left(\sum_{cycl}x^2\right)^2}{\displaystyle \sum_{cycl}x^2}\\ &=\sum_{cycl}x^2. \end{align}

### Acknowledgment

CutTheKnotMath facebook page. The problem was originally published at the Romanian Mathematical Magazine. Solution 1 is by Diego Alvariz; Solution 2 is by Soumava Chakraborty.