# An Inequality with Altitudes and Medians

### Solution

Since $\Delta ABC\,$ is acute, we may assume $a=\sqrt{y+z},\,$ $b=\sqrt{z+x},\,$ and $c=\sqrt{x+y},\,$ with $x,y,z\gt 0.$

WLOG, x\le y\le z,\,$implying$a\ge b\ge c,\,h_c=\max (h_a,h_b,h_c)\,$and$m_a=\min (m_a,m_b,m_c).\,$By Heron's formula,$2[\Delta ABC]=\sqrt{xy+yz+zx}.\,$Also,$2[\Delta ABC=h_cc=h_c\sqrt{x+y},\,$i.e.,$\sqrt{xy+yz+zx}=h_c\sqrt{x+y},\,$making$\displaystyle h_c\frac{\sqrt{xy+yz+zx}}{\sqrt{x+y}}.\,$The left inequality is equivalent to$\displaystyle 2\sqrt{\frac{3(xy+yz+zx)}{x+y}}\ge\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}.$By Jensen's inequality for$\sqrt{t}\,$on$(0,\infty ),\,\sqrt{6(x+y+z)}\ge\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}.$Suffice it to show that$\displaystyle 2\sqrt{\frac{3(xy+yz+zx)}{x+y}}\ge\sqrt{6(x+y+z)}.$This is equivalent to$2(xy+yz+zx)\ge (x+y)(x+y+z),\,$i.e., to$z(x+y)\ge x^2+y^2.\,$But$z(x+y)\ge y(x+y)=xy+y^2\ge x^2+y^2.$For the right inequality, recollect that 2m_a=\sqrt{4x+y+z}.\,$ So the right inequality is equivalent to

$\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\ge\sqrt{3(4x+y+z)}.$

$\displaystyle 2\sum_{cycl}\sqrt{(x+y)(z+x)}\ge (x+y)+(z+x)+8x.$
But $\sqrt{(x+y)(y+z)}\ge (x+y),\,$ $\sqrt{(z+x)(y+z)}\ge (z+x),\,$ $2\sqrt{(x+y)(z+x)}\ge 4x,\,$ $\sqrt{(x+y)(y+z)}\ge 2x,\,$ and \sqrt{(z+x)(y+z)}\ge 2x.\,\$ Adding up completes the proof.