Question Number 194325 by horsebrand11 last updated on 03/Jul/23
$$\:\:\cancel{\mathcal{X}} \\ $$
Answered by AST last updated on 04/Jul/23
$$\sqrt[{\mathrm{4}_{} }]{\frac{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} }{\mathrm{3}}}\geqslant\sqrt{\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{\mathrm{3}}}=\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\Rightarrow\Sigma{x}^{\mathrm{4}} \geqslant\frac{\mathrm{25}}{\mathrm{3}} \\ $$$$\Sigma{x}^{\mathrm{4}} =\left(\Sigma{x}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\Sigma{x}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{25}−\mathrm{2}\Sigma{x}^{\mathrm{2}} {y}^{\mathrm{2}} \\ $$$$\Rightarrow\Sigma{x}^{\mathrm{4}} \leqslant\mathrm{25}−\mathrm{0}\left({since}\:\Sigma{x}^{\mathrm{2}} {y}^{\mathrm{2}} \geqslant\mathrm{0}\:{and}\:{max}\:{occurs}\:{at}\:\mathrm{0}\right) \\ $$