Question Number 196938 by sonukgindia last updated on 04/Sep/23
Answered by AST last updated on 04/Sep/23
$${xy}+{yz}+{zx}\leqslant\mathrm{147}\left({equality}\:{when}\:{x}={y}={z}=\underset{−} {+}\mathrm{7}\right) \\ $$$$\left(\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{\mathrm{3}}\right)\geqslant\left(\frac{{x}+{y}+{z}}{\mathrm{3}}\right)^{\mathrm{2}} \Rightarrow\mathrm{3}\left(\mathrm{147}\right)\geqslant\left({x}+{y}+{z}\right)^{\mathrm{2}} \\ $$$$\Rightarrow−\mathrm{21}\leqslant\left({x}+{y}+{z}\right)\leqslant\mathrm{21} \\ $$$$\Rightarrow{x}+{y}+{z}−{xy}−{yz}−{zx}\geqslant−\mathrm{147}−\mathrm{21}=−\mathrm{168} \\ $$$$\left({equality}\:{when}\:{x}={y}={z}=−\mathrm{7}\right) \\ $$