## Monday, September 14, 2015

### Hard Inequality Problem

Let $x,\,y,\,z$ be real numbers such that $6x+2y+3z=12+xyz$.

Prove that $(x^2+1)(y^2+9)(z^2+4)\ge 144$.

Note that

$(x^2+1)(y^2+9)(z^2+4)-(6x+2y+3z-xyz)^2$

$=4 x^2 y^2+12 x^2 y z+9 x^2 z^2+4 x y^2 z+6 x y z^2-24 x y-36 x z+y^2 z^2-12 y z+36$

$\small=(9 x^2 z^2+y^2 z^2+6 x y z^2)+(4 x^2 y^2-24 x y+36)+(12 x^2 y z+4 x y^2 z-36 x z-12 y z)$

$=(3xz+yz)^2+4(xy-3)^2+4(3 x^2 y z+x y^2 z-9 x z-3 y z)$

$=(3xz+yz)^2+4(xy-3)^2+4(3 x^2 y z-9 x z+x y^2 z-3 y z)$

$=(3xz+yz)^2+4(xy-3)^2+4(3xz(xy-3)+yz(xy-3))$

$=(3xz+yz)^2+4(3xz+yz)(xy-3)+4(xy-3)^2$

[MATH]\color{yellow}\bbox[5px,purple]{=a^2+4ab+4b^2=(a+2b)^2}[/MATH]

$=(3xz+yz+2(xy-3))^2$

And since we're given $6x+2y+3z=12+xyz$, i.e. $6x+2y+3z-xyz=12$, we can conclude that

$\small(x^2+1)(y^2+9)(z^2+4)=(6x+2y+3z-xyz)^2+(3xz+yz+2(xy-3))^2\ge 12^2=144$

and we're hence done.