Question Number 192612 by York12 last updated on 22/May/23
$$\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{y}}\:+\:\frac{\mathrm{1}}{{z}}\:=\:\mathrm{1}\: \\ $$$${find}\:{the}\:{minimum}\:{value}\:{of}\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \\ $$
Commented by Frix last updated on 23/May/23
$$\mathrm{Yes}\:\mathrm{of}\:\mathrm{course}! \\ $$
Commented by deleteduser1 last updated on 23/May/23
$${For}\:{positive}\:{x},{y},{z};\:{min}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)=\mathrm{27}\:{at}\: \\ $$$${x}={y}={z}=\mathrm{3} \\ $$
Commented by York12 last updated on 23/May/23
$${exactly} \\ $$
Answered by Subhi last updated on 23/May/23
$$ \\ $$$${Apply}\:{titu}'{s}\:{Lemma}\:{enquality} \\ $$$$\mathrm{1}=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\geqslant\frac{\mathrm{9}}{{x}+{y}+{z}}\geqslant\frac{\mathrm{9}}{\:\sqrt{\mathrm{3}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)}} \\ $$$$\sqrt{\mathrm{3}\left({x}^{{x}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)}\geqslant\mathrm{9} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \geqslant\frac{\mathrm{81}}{\mathrm{3}}=\mathrm{27} \\ $$$${then},\:{the}\:{mini}\:{value}\:{is}\:\mathrm{27} \\ $$