# Leo Giugiuc's Inequality for the Medians

### Solution 1

\begin{align} 2m_a &= \sqrt{2b^2+2c^2-a^2}=\sqrt{(b+c)^2-a^2+(b-c)^2}\\ &=\sqrt{(b+c-a)(b=c+a)-(b-c)^2}\\ &=\sqrt{4s(s-a)+(b-c)^2}. \end{align}

Similarly, $2m_b=\sqrt{4s(s-b)+(c-a)^2}\,$ and $2m_c=\sqrt{4s(s-c)+(a-b)^2}.\,$ But $\sqrt{x}+\sqrt{y}+\sqrt{z}\le \sqrt{3(x+y+z)},\,$ such that

$2(m_a+m_b+m_c)\le\sqrt{12s^2+3[(a-b)^2+(b-c)^2+(c-a)^2]},$

which is equivalent to the required inequality.

### Solution 2

Squaring the left-hand side, using the formula for the square of a median, we estimate

\begin{align} \left(\sum_{cycl}m_a\right)^2&=\sum_{cycl}m_a^2+\sum_{cycl}(2m_am_b)\\ &\le\sum_{cycl}m_a^2+\sum_{cycl}(m_a^2+m_b^2)=3\sum_{cycl}m_a^2\\ &=3\cdot\frac{3}{4}\sum_{cycl}a^2. \end{align}

For the right-hand side,

\displaystyle\begin{align} 3s^2+\frac{3}{4}\sum_{cycl}(a-b)^2&=\frac{3}{4}\left[\left(\sum_{cycl}a\right)^2+\sum_{cycl}(a-b)^2\right]\\ &=\frac{3}{4}\left[\sum_{cycl}a^2+2\sum_{cycl}ab+2\sum_{cycl}a^2-2\sum_{cycl}ab\right]\\ &=\frac{9}{4}\sum_{cycl}a^2, \end{align}

which proves the inequality.

### Acknowledgment

Leo Giugiuc has kindly posted at the CutTheKnotMath facebook page his problem and solution (Solution 1).