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Question Number 194325 by horsebrand11 last updated on 03/Jul/23

$$\cancel{\mathcal{X}}$$

Answered by deleteduser1 last updated on 04/Jul/23

$$\sqrt[4_{}]{\frac{x^4+y^4+z^4}{3}}⩾\sqrt{\frac{x^2+y^2+z^2}{3}}=\sqrt{\frac{5}{3}}\Rightarrow \mathrm{\Sigma }{x}^4⩾\frac{25}{3}$$$$\mathrm{\Sigma }{x}^4={\left(\mathrm{\Sigma }{x}^2\right)}^2-2\mathrm{\Sigma }{x}^2{y}^2=25-2\mathrm{\Sigma }{x}^2{y}^2$$$$\Rightarrow \mathrm{\Sigma }{x}^4⩽25-0(since\mathrm{\Sigma }{x}^2{y}^2⩾0andmaxoccursat0)$$

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