A Friend's Geometric Inequality(or is it?)

by XmL, Aug 20, 2014, 4:37 AM

If $a,b,c$ are the sides of a triangle, prove the inequality:
$\frac {a^2+b^2+c^2}{16}\Sigma \frac {a^2}{M_b^2M_c^2}\ge 1$ .

Applying the median length formula should be the obvious first step:

LHS= $\frac {a^2+b^2+c^2}{16}\Sigma \frac {a^2}{(\frac {a^2+b^2}{2}-\frac {c^2}{4})(\frac {a^2+c^2}{2}-\frac {b^2}{4})}$

$=(a^2+b^2+c^2)\Sigma \frac {a^2}{(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)}$

To make things easier we apply substitution $x=a^2,y=b^2,z=c^2$ :

$=(x+y+z)\Sigma \frac {x}{(2x+2y-z)(2x+2z-y)}$

Since the expression is homogenous, we can let $x+y+z=1$ , so the left-hand side becomes:

$=\Sigma \frac {x}{(2-3z)(2-3y)}=\frac {-3\Sigma x^2+2\Sigma x}{(2-3x)(2-3y)(2-3z)}=\frac {-3\Sigma x^2+2}{-4+18\Sigma xy-27xyz}$

Recall that $(2-3x)(2-3y)(2-3z)>0$ since the terms each represents a length, so after some expansion and rearranging the inequality is equivalent to:

$9xyz+2\ge 6\Sigma xy+\Sigma x^2=4\Sigma xy+1$

$\Leftrightarrow 9xyz+(x+y+z)^3\ge 4(x+y+z)\Sigma xy$

$\Leftrightarrow x^3+y^3+z^3+3xyz\ge \Sigma x^2y$

which is true by schur's. $_\Box$ .

This post has been edited 2 times. Last edited by XmL, Aug 20, 2014, 7:01 AM

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Good blog! I hope to see more posts here!
by WizardMath, Feb 3, 2018, 6:57 PM
Thanks Harry Potter
by XmL, Jul 30, 2014, 10:47 PM
Hey dude. Nice blog. I like your work.
by gamjawon, Jul 30, 2014, 7:43 AM
Thanks...although most of the posts here won't be about geometry...rightnow I'm trying to improve my skills in other areas such as number theory and combo.
by XmL, Jun 5, 2014, 1:11 AM
Hi Geometer,,nice blog..
by vanu1996, Jun 4, 2014, 1:27 PM
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by zabihpourmahdi, Jan 11, 2013, 5:34 PM

6 shouts

Geometry Treasures

A Friend's Geometric Inequality(or is it?)

by XmL, Aug 20, 2014, 4:37 AM

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