An Inequality with Altitudes and Medians

Problem

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

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

Squaring leads to

$\displaystyle 2\sum_{cycl}\sqrt{(x+y)(z+x)}\ge (x+y)+(z+x)+8x.$

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.

Acknowledgment

The problem and its solution have been kindly communicated to me by Leo Giugiuc. The problem is by R. Satnoianu; it has been published at the Romanian Mathematical Magazine.

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