Question Number 218005 by jacklau last updated on 25/Mar/25
$$\: \\ $$$$\: \\ $$$$\:\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} +\boldsymbol{\mathrm{z}}^{\mathrm{2}} =\mathrm{1},\: \\ $$$$\:\boldsymbol{\mathrm{maximum}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{x}}\mathrm{y}+\mathrm{2yz}\:=\:… \\ $$
Commented by Ghisom last updated on 26/Mar/25
$$\mathrm{I}\:\mathrm{get} \\ $$$$−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\leqslant\left({x}+\mathrm{2}{z}\right){y}\leqslant\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 26/Mar/25
$${yes}.\:{this}\:{is}\:{correct}.\:{maximum}\:{is}\:{at} \\ $$$${x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}},\:{y}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}},\:{z}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{10}}} \\ $$