Question

Show that
(1) ${(x+y+z)}^3>27(y+z-x)(z+x-y)(x+y-z)$
(2) $xyz>(y+z-x)(z+x-y)(x+y-z)$.

Answer 1

According to AM -GM Inequality,

$AM\ge GM$

That is arithematic mean is greater than equal to geometric mean.

$\frac{(y+z-x)+(z+x-y)+(x+y-z)}{3}>{\left((y+z-x)(z+x-y)(x+y-z)\right)}^{\frac{1}{3}}$

$\frac{(x+y+z)}{3}>{\left((y+z-x)(z+x-y)(x+y-z)\right)}^{\frac{1}{3}}$

${\left(\frac{(x+y+z)}{3}\right)}^3>\left((y+z-x)(z+x-y)(x+y-z)\right)$

$\frac{{(x+y+z)}^3}{27}>\left((y+z-x)(z+x-y)(x+y-z)\right)$

${(x+y+z)}^3>27(y+z-x)(z+x-y)(x+y-z)$

$\frac{(y+z-x)+(z+x-y)}{2}>\sqrt{(y+z-x)(z+x-y)}$

$\frac{(y+z-x)+(y+x-z)}{2}>\sqrt{(y+z-x)(y+x-z)}$

$\frac{(y+x-z)+(z+x-y)}{2}>\sqrt{(y+x-z)(z+x-y)}$

$\frac{2z}{2}>\sqrt{(y+z-x)(z+x-y)}$

$z>\sqrt{(y+z-x)(z+x-y)}$

$y>\sqrt{(y+z-x)(y+x-z)}$

$x>\sqrt{(y+x-z)(z+x-y)}$

Multiply the above three equation, we get

$xyz>\sqrt{{(y+z-x)}^2{(z+x-y)}^2{(x+y-z)}^2}$

$xyz>(y+z-x)(z+x-y)(x+y-z)$