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Question Number 76904 by mpsicasa last updated on 31/Dec/19
montrer que: ∀x,y,z>0 x^3 +2y^2 +4z≥6xy^(2/3) z^(1/3)
$$\mathrm{montrer}\:\mathrm{que}: \\ $$$$\forall{x},{y},{z}>\mathrm{0}\:\:{x}^{\mathrm{3}} +\mathrm{2}{y}^{\mathrm{2}} +\mathrm{4}{z}\geqslant\mathrm{6}{xy}^{\mathrm{2}/\mathrm{3}} {z}^{\mathrm{1}/\mathrm{3}} \\ $$
Commented by mr W last updated on 31/Dec/19
A.M.≥G.M. ((a+b+c)/3)≥((abc))^(1/3) a+b+c≥3((abc))^(1/3) x^3 +2y^2 +4z≥3((x^3 ×2y^2 ×4z))^(1/3) =6xy^(2/3) z^(1/3)
$${A}.{M}.\geqslant{G}.{M}. \\ $$$$\frac{{a}+{b}+{c}}{\mathrm{3}}\geqslant\sqrt[{\mathrm{3}}]{{abc}} \\ $$$${a}+{b}+{c}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}} \\ $$$${x}^{\mathrm{3}} +\mathrm{2}{y}^{\mathrm{2}} +\mathrm{4}{z}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} ×\mathrm{2}{y}^{\mathrm{2}} ×\mathrm{4}{z}}=\mathrm{6}{xy}^{\mathrm{2}/\mathrm{3}} {z}^{\mathrm{1}/\mathrm{3}} \\ $$
Commented by Zainal Arifin last updated on 13/Apr/20
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Answered by MJS last updated on 31/Dec/19
f(x, y, z)=x^3 +2y^2 +4z−6xy^(2/3) z^(1/3) searching for the minimum (df/dx)=3x^2 −6y^(2/3) z^(1/3) =0 ⇒ x=(√2)y^(1/3) z^(1/6) ⇒ f((√2)y^(1/3) z^(1/6) , y, z)=2y^2 +4z−4(√2)yz^(1/2) (df/dy)=4y−4(√2)z^(1/2) =0 ⇒ y=(√2)z^(1/2) ⇒ x=(4)^(1/3) z^(1/3) ⇒ f((4)^(1/3) z^(1/3) , (√2)z^(1/2) , z)=0 ⇒ 0=minimum of f(x, y, z) with x, y, z >0 ⇒ x^3 +2y^2 +4z−6xy^(2/3) z^(1/3) ≥0 x^3 +2y^2 +4z≥6xy^(2/3) z^(1/3)
$${f}\left({x},\:{y},\:{z}\right)={x}^{\mathrm{3}} +\mathrm{2}{y}^{\mathrm{2}} +\mathrm{4}{z}−\mathrm{6}{xy}^{\frac{\mathrm{2}}{\mathrm{3}}} {z}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\mathrm{searching}\:\mathrm{for}\:\mathrm{the}\:\mathrm{minimum} \\ $$$$\frac{{df}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{6}{y}^{\frac{\mathrm{2}}{\mathrm{3}}} {z}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{0}\:\Rightarrow\:{x}=\sqrt{\mathrm{2}}{y}^{\frac{\mathrm{1}}{\mathrm{3}}} {z}^{\frac{\mathrm{1}}{\mathrm{6}}} \\ $$$$\Rightarrow \\ $$$${f}\left(\sqrt{\mathrm{2}}{y}^{\frac{\mathrm{1}}{\mathrm{3}}} {z}^{\frac{\mathrm{1}}{\mathrm{6}}} ,\:{y},\:{z}\right)=\mathrm{2}{y}^{\mathrm{2}} +\mathrm{4}{z}−\mathrm{4}\sqrt{\mathrm{2}}{yz}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\frac{{df}}{{dy}}=\mathrm{4}{y}−\mathrm{4}\sqrt{\mathrm{2}}{z}^{\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{0}\:\Rightarrow\:{y}=\sqrt{\mathrm{2}}{z}^{\frac{\mathrm{1}}{\mathrm{2}}} \:\Rightarrow\:{x}=\sqrt[{\mathrm{3}}]{\mathrm{4}}{z}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\Rightarrow \\ $$$${f}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}{z}^{\frac{\mathrm{1}}{\mathrm{3}}} ,\:\sqrt{\mathrm{2}}{z}^{\frac{\mathrm{1}}{\mathrm{2}}} ,\:{z}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{0}=\mathrm{minimum}\:\mathrm{of}\:{f}\left({x},\:{y},\:{z}\right)\:\mathrm{with}\:{x},\:{y},\:{z}\:>\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{3}} +\mathrm{2}{y}^{\mathrm{2}} +\mathrm{4}{z}−\mathrm{6}{xy}^{\frac{\mathrm{2}}{\mathrm{3}}} {z}^{\frac{\mathrm{1}}{\mathrm{3}}} \geqslant\mathrm{0} \\ $$$${x}^{\mathrm{3}} +\mathrm{2}{y}^{\mathrm{2}} +\mathrm{4}{z}\geqslant\mathrm{6}{xy}^{\frac{\mathrm{2}}{\mathrm{3}}} {z}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$

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