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Question Number 141405 by iloveisrael last updated on 18/May/21

$$Findtheminimumvalueofk$$ $$suchthatforarbitrarya,b>0$$ $$wehave\sqrt[3]{a}+\sqrt[3]{b}⩽k\sqrt[3]{a+b}$$

Answered by EDWIN88 last updated on 18/May/21

$$\mathrm{consider}\mathrm{the}\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)=\sqrt[3]{\mathrm{x}}$$ $$\mathrm{we}\mathrm{have}\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\frac{1}{3}{\mathrm{x}}^{-\frac{2}{3}}\mathrm{\&}\mathrm{f}{}^{″}\left(\mathrm{x}\right)=-\frac{2}{9}{\mathrm{x}}^{-\frac{5}{3}}$$ $$\mathrm{for}\mathrm{any}\mathrm{x}\in (0,\mathrm{\infty })\mathrm{thus}\mathrm{f}\left(\mathrm{x}\right)\mathrm{concave}\mathrm{on}\mathrm{the}$$ $$\mathrm{interval}(0,\mathrm{\infty }).\mathrm{By}{\mathrm{Jensen}}^{\prime }\mathrm{s}\mathrm{inequality}$$ $$\mathrm{we}\mathrm{deduce}$$ $$\frac{1}{2}\mathrm{f}\left(\mathrm{a}\right)+\frac{1}{2}\mathrm{f}\left(\mathrm{b}\right)⩽\mathrm{f}\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)$$ $$\frac{\sqrt[3]{\mathrm{a}}+\sqrt[3]{\mathrm{b}}}{2}⩽\sqrt[3]{\frac{\mathrm{a}+\mathrm{b}}{2}}$$ $$\sqrt[3]{\mathrm{a}}+\sqrt[3]{\mathrm{b}}⩽\frac{2\sqrt[3]{\mathrm{a}+\mathrm{b}}}{\sqrt[3]{2}}=\frac{2\sqrt[3]{4}}{2}(\mathrm{a}+\mathrm{b})$$ $$\mathrm{minimum}\mathrm{value}\mathrm{of}k\mathrm{is}\sqrt[3]{4}\divideontimes$$

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