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Question Number 205904 by RoseAli last updated on 02/Apr/24

Answered by mr W last updated on 02/Apr/24

$$a+b+c=2$$$$\sqrt[3]{abc}⩽\frac{a+b+c}{3}=\frac{2}{3}$$$$abc={\left(\sqrt[3]{abc}\right)}^3⩽{\left(\frac{2}{3}\right)}^3=\frac{8}{27}$$$$(1+\frac{2}{a})(1+\frac{2}{b})(1+\frac{2}{c})$$$$=1+2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+4(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})+\frac{8}{abc}$$$$=1+2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+4\left(\frac{a+b+c}{abc}\right)+\frac{8}{abc}$$$$=1+2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+\frac{8}{abc}+\frac{8}{abc}$$$$=1+2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+\frac{16}{abc}$$$$⩾1+2\times \frac{3}{\sqrt[3]{abc}}+\frac{16}{abc}$$$$⩾1+6\times \frac{3}{2}+16\times \frac{27}{8}=64$$$$i.e.minimumof(1+\frac{2}{a})(1+\frac{2}{b})(1+\frac{2}{c})$$$$is64.$$

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