let-x-y-z-gt-0-such-that-x-4-y-4-z-4-x-2-y-2-z-2-find-the-minimum-of-the-expression-P-x-2-y-y-2-z-z-2-x-

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

let x;y;z > 0 such that x^4 +y^4 +z^4 = x^2 +y^2 +z^2 find the minimum of the expression: P = (x^2 /y) + (y^2 /z) + (z^2 /x)

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

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) But x; y; z > 0 ⇒ (x; y; z) = (1; 1; 1) P = (x^2 /y) + (y^2 /z) + (z^2 /x) = (1^2 /1) + (1^2 /1) + (1^2 /1) = 1 + 1 + + 1 = 3 P = 3

$${x}^{\mathrm{4}} \:+\:{y}^{\mathrm{4}} \:+\:{z}^{\mathrm{4}} \:=\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \\ $$$$\left({x}_{\mathrm{1}} ;\:{y}_{\mathrm{1}} ;\:{z}_{\mathrm{1}} \right)\:=\:\left(−\mathrm{1};\:−\mathrm{1};\:−\mathrm{1}\right) \\ $$$$\left({x}_{\mathrm{2}} ;\:{y}_{\mathrm{2}} ;\:{z}_{\mathrm{2}} \right)\:=\:\left(\mathrm{0};\:\mathrm{0};\:\mathrm{0}\right) \\ $$$$\left({x}_{\mathrm{3}} ;\:{y}_{\mathrm{3}} ;\:{z}_{\mathrm{3}} \right)\:=\:\left(\mathrm{1};\:\mathrm{1};\:\mathrm{1}\right) \\ $$$$\mathrm{But}\:{x};\:{y};\:{z}\:>\:\mathrm{0} \\ $$$$\Rightarrow\:\left({x};\:{y};\:{z}\right)\:=\:\left(\mathrm{1};\:\mathrm{1};\:\mathrm{1}\right) \\ $$$$\mathrm{P}\:=\:\frac{{x}^{\mathrm{2}} }{{y}}\:+\:\frac{{y}^{\mathrm{2}} }{{z}}\:+\:\frac{{z}^{\mathrm{2}} }{{x}}\:=\:\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{1}}\:+\:\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{1}}\:+\:\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{1}}\:=\:\mathrm{1}\:+\:\mathrm{1}\:+ \\ $$$$+\:\mathrm{1}\:=\:\mathrm{3} \\ $$$$\mathrm{P}\:=\:\mathrm{3} \\ $$

Commented by HongKing last updated on 01/Jan/22

My dear Sir thank you, but need to prove P≥3 for all x;y;z such that x^4 +y^4 +z^4 = x^2 +y^2 +z^2

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

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