Question Number 55150 by naka3546 last updated on 18/Feb/19
Answered by tanmay.chaudhury50@gmail.com last updated on 18/Feb/19
$$\frac{\left(\mathrm{1}+{x}\right)+\left(\mathrm{1}+{y}\right)+\left(\mathrm{1}+{z}\right)}{\mathrm{3}}\geqslant\sqrt[{\mathrm{3}}]{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{y}\right)\left(\mathrm{1}+{z}\right)}\:\geqslant\frac{\mathrm{3}}{\frac{\mathrm{1}}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{1}+{y}}+\frac{\mathrm{1}}{\mathrm{1}+{z}}} \\ $$$$\mathrm{1}+\frac{{x}+{y}+{z}}{\mathrm{3}}\geqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\frac{{x}+{y}+{z}}{\mathrm{3}}\geqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}+{y}+{z}\geqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${now} \\ $$$$\frac{\mathrm{1}+\mathrm{4}{x}+\mathrm{1}+\mathrm{4}{y}+\mathrm{1}+\mathrm{4}{z}}{\mathrm{3}}\geqslant\frac{\mathrm{3}}{\frac{\mathrm{1}}{\mathrm{1}+\mathrm{4}{x}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{4}{y}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{4}{z}}} \\ $$$$\mathrm{1}+\frac{\mathrm{4}}{\mathrm{3}}\left({x}+{y}+{z}\right)\geqslant\frac{\mathrm{3}}{{k}} \\ $$$$\mathrm{1}+\frac{\mathrm{4}}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{2}}\geqslant\frac{\mathrm{3}}{{k}} \\ $$$$\mathrm{3}\geqslant\frac{\mathrm{3}}{{k}} \\ $$$$\mathrm{3}{k}\geqslant\mathrm{3} \\ $$$${so}\:{minimum}\:{value}\:{of}\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{4}{x}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{4}{y}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{4}{z}}=\mathrm{1} \\ $$
Answered by tm888 last updated on 19/Feb/19
