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Question Number 162788 by HongKing last updated on 01/Jan/22

$$\mathrm{let}\mathrm{x};\mathrm{y};\mathrm{z}>0$$$$\mathrm{such}\mathrm{that}{\mathrm{x}}^4+{\mathrm{y}}^4+{\mathrm{z}}^4={\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2$$$$\mathrm{find}\mathrm{the}\mathrm{minimum}\mathrm{of}\mathrm{the}\mathrm{expression}:$$$$\mathrm{P}=\frac{{\mathrm{x}}^2}{\mathrm{y}}+\frac{{\mathrm{y}}^2}{\mathrm{z}}+\frac{{\mathrm{z}}^2}{\mathrm{x}}$$

Answered by alephzero last updated on 01/Jan/22

$$x^4+y^4+z^4=x^2+y^2+z^2$$$$(x_1;y_1;z_1)=(-1;-1;-1)$$$$(x_2;y_2;z_2)=(0;0;0)$$$$(x_3;y_3;z_3)=(1;1;1)$$$$\mathrm{But}x;y;z>0$$$$\Rightarrow (x;y;z)=(1;1;1)$$$$\mathrm{P}=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}=\frac{1^2}{1}+\frac{1^2}{1}+\frac{1^2}{1}=1+1+$$$$+1=3$$$$\mathrm{P}=3$$

Commented by HongKing last updated on 01/Jan/22

$$\mathrm{My}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you},\mathrm{but}$$$$\mathrm{need}\mathrm{to}\mathrm{prove}\mathrm{P}⩾3\mathrm{for}\mathrm{all}\mathrm{x};\mathrm{y};\mathrm{z}$$$$\mathrm{such}\mathrm{that}{\mathrm{x}}^4+{\mathrm{y}}^4+{\mathrm{z}}^4={\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2$$

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