# ​if x,y,z are distinct positive numbers ,prove that (x+y)(y+z)(z+x)>8xyz.further if x+y+z=1 show that (1-x)(1-y)(1-z)>8xyz

Dear student

For positive numbers, Arithmetic mean is always greater than or equal to geometric mean.

$\frac{\mathrm{x}+\mathrm{y}}{2}\ge \sqrt{\mathrm{xy}}...\left(1\right)\phantom{\rule{0ex}{0ex}}\frac{\mathrm{y}+\mathrm{z}}{2}\ge \sqrt{\mathrm{yz}}...\left(2\right)\phantom{\rule{0ex}{0ex}}\frac{\mathrm{x}+\mathrm{z}}{2}\ge \sqrt{\mathrm{xz}}...\left(3\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Multiplying}\mathrm{the}\mathrm{three}\mathrm{equations},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)\left(\frac{\mathrm{y}+\mathrm{z}}{2}\right)\left(\frac{\mathrm{x}+\mathrm{z}}{2}\right)\ge \sqrt{\mathrm{xy}}\sqrt{\mathrm{yz}}\sqrt{\mathrm{xz}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{y}+\mathrm{z}\right)\left(\mathrm{z}+\mathrm{x}\right)\ge 8\mathrm{xyz}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{If}\mathrm{x}+\mathrm{y}+\mathrm{z}=1,\mathrm{then}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{x}+\mathrm{y}=1-\mathrm{z}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{z}+\mathrm{y}=1-\mathrm{x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{x}+\mathrm{z}=1-\mathrm{y}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{y}+\mathrm{z}\right)\left(\mathrm{z}+\mathrm{x}\right)\ge 8\mathrm{xyz}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\left(1-\mathrm{z}\right)\left(1-\mathrm{x}\right)\left(1-\mathrm{y}\right)\ge 8\mathrm{xyz}\phantom{\rule{0ex}{0ex}}$

Regards

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