Question Number 144600 by mathdanisur last updated on 26/Jun/21
$${if}\:{x},{y},{z}>\mathrm{0}\:;\:{xy}+{yz}+{zx}=\mathrm{1}\:{prove}\:{that}: \\ $$ $${xyz}\:+\:\sqrt[{\mathrm{3}}]{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\left(\mathrm{1}+{y}^{\mathrm{3}} \right)\left(\mathrm{1}+{z}^{\mathrm{3}} \right)}\:\geqslant\:\mathrm{1} \\ $$
Answered by mindispower last updated on 27/Jun/21
$${x},{y},{z}>\mathrm{0} \\ $$ $${or}\:{only}\:{z}>\mathrm{0} \\ $$
Commented bymathdanisur last updated on 27/Jun/21
$${Thanks}\:{Sir},\:{how}... \\ $$