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Question Number 145383 by mathdanisur last updated on 04/Jul/21

$$ifa;b;c\in {\mathbb{R}}^{+}$$$$find{(\sqrt[3]{abc}+\frac{1}{a}+\frac{1}{2b}+\frac{1}{4c})}_{\mathit{m}\mathit{i}\mathit{n}}=?$$

Answered by mnjuly1970 last updated on 04/Jul/21

$$\mathrm{A}⩾\sqrt[3]{abc}+3\sqrt[3]{\frac{1}{8abc}}$$$$=\sqrt[3]{abc}+\frac{3}{2}\sqrt[3]{\frac{1}{abc}}\stackrel{am-gm}{⩾}2\sqrt{\frac{3}{2}}$$$${\mathrm{A}}_{min}=2\sqrt{\frac{3}{2}}=\sqrt{6}$$

Commented by mathdanisur last updated on 04/Jul/21

$$ThanksSir,answer\sqrt{6}$$

Commented by mnjuly1970 last updated on 04/Jul/21

$$thankyousomuchand$$$$yourmention...$$

Commented by mathdanisur last updated on 04/Jul/21

$$coolSer,thankyou$$

Answered by ajfour last updated on 04/Jul/21

$$let\frac{1}{a}=x,\frac{1}{2b}=y,\frac{1}{4c}=z$$$$\Rightarrow f(x,y,z)=\frac{1/2}{{\left(xyz\right)}^{1/3}}+x+y+z$$$$minimumshouldofcourse$$$$bewhenx=y=z\left(symmetry\right)$$$$\Rightarrow f_{min}=\frac{1}{2x}+3x$$$$where\frac{1}{2x}=3x\Rightarrow x^2=\frac{1}{6}$$$$f_{min}=6x=\frac{6}{\sqrt{6}}=\sqrt{6}.$$

Commented by mathdanisur last updated on 04/Jul/21

$$coolSer,thankyou$$

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