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Question-205904




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  ((abc))^(1/3) ≤((a+b+c)/3)=(2/3)  abc=(((abc))^(1/3) )^3 ≤((2/3))^3 =(8/(27))  (1+(2/a))(1+(2/b))(1+(2/c))  =1+2((1/a)+(1/b)+(1/c))+4((1/(ab))+(1/(bc))+(1/(ca)))+(8/(abc))  =1+2((1/a)+(1/b)+(1/c))+4(((a+b+c)/(abc)))+(8/(abc))  =1+2((1/a)+(1/b)+(1/c))+(8/(abc))+(8/(abc))  =1+2((1/a)+(1/b)+(1/c))+((16)/(abc))  ≥1+2×(3/( ((abc))^(1/3) ))+((16)/(abc))  ≥1+6×(3/2)+16×((27)/8)=64  i.e. minimum of (1+(2/a))(1+(2/b))(1+(2/c))  is 64.
$${a}+{b}+{c}=\mathrm{2} \\ $$$$\sqrt[{\mathrm{3}}]{{abc}}\leqslant\frac{{a}+{b}+{c}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${abc}=\left(\sqrt[{\mathrm{3}}]{{abc}}\right)^{\mathrm{3}} \leqslant\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} =\frac{\mathrm{8}}{\mathrm{27}} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{2}}{{a}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{b}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{c}}\right) \\ $$$$=\mathrm{1}+\mathrm{2}\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)+\mathrm{4}\left(\frac{\mathrm{1}}{{ab}}+\frac{\mathrm{1}}{{bc}}+\frac{\mathrm{1}}{{ca}}\right)+\frac{\mathrm{8}}{{abc}} \\ $$$$=\mathrm{1}+\mathrm{2}\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)+\mathrm{4}\left(\frac{{a}+{b}+{c}}{{abc}}\right)+\frac{\mathrm{8}}{{abc}} \\ $$$$=\mathrm{1}+\mathrm{2}\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)+\frac{\mathrm{8}}{{abc}}+\frac{\mathrm{8}}{{abc}} \\ $$$$=\mathrm{1}+\mathrm{2}\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)+\frac{\mathrm{16}}{{abc}} \\ $$$$\geqslant\mathrm{1}+\mathrm{2}×\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{{abc}}}+\frac{\mathrm{16}}{{abc}} \\ $$$$\geqslant\mathrm{1}+\mathrm{6}×\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{16}×\frac{\mathrm{27}}{\mathrm{8}}=\mathrm{64} \\ $$$${i}.{e}.\:{minimum}\:{of}\:\left(\mathrm{1}+\frac{\mathrm{2}}{{a}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{b}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{c}}\right) \\ $$$${is}\:\mathrm{64}. \\ $$

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