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Question Number 189124 by Rupesh123 last updated on 12/Mar/23

Answered by witcher3 last updated on 15/Mar/23

x+y+x≥3((xyz))^(1/3)   ⇔xyz≥3((xyz))^(1/3)  ;let a=xyz  ⇔a≥3(a)^(1/3) ⇒a^(2/3) ≥3⇒a≥((27))^(1/2) =3(√3)  a+(1/a)=f(a)≥2,∀a>0  f′(a)=1−(1/a^2 ),∀a>1 f is increase  f(3(√(3)))=(√(27))+(1/( (√(27))))=((28)/( (√(27))))=((28)/(3(√3)))

$$\mathrm{x}+\mathrm{y}+\mathrm{x}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{xyz}} \\ $$$$\Leftrightarrow\mathrm{xyz}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{xyz}}\:;\mathrm{let}\:\mathrm{a}=\mathrm{xyz} \\ $$$$\Leftrightarrow\mathrm{a}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{a}}\Rightarrow\mathrm{a}^{\frac{\mathrm{2}}{\mathrm{3}}} \geqslant\mathrm{3}\Rightarrow\mathrm{a}\geqslant\sqrt[{\mathrm{2}}]{\mathrm{27}}=\mathrm{3}\sqrt{\mathrm{3}} \\ $$$$\mathrm{a}+\frac{\mathrm{1}}{\mathrm{a}}=\mathrm{f}\left(\mathrm{a}\right)\geqslant\mathrm{2},\forall\mathrm{a}>\mathrm{0} \\ $$$$\mathrm{f}'\left(\mathrm{a}\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} },\forall\mathrm{a}>\mathrm{1}\:\mathrm{f}\:\mathrm{is}\:\mathrm{increase} \\ $$$$\mathrm{f}\left(\mathrm{3}\sqrt{\left.\mathrm{3}\right)}=\sqrt{\mathrm{27}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{27}}}=\frac{\mathrm{28}}{\:\sqrt{\mathrm{27}}}=\frac{\mathrm{28}}{\mathrm{3}\sqrt{\mathrm{3}}}\right. \\ $$

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