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Question-64186




Question Number 64186 by Hope last updated on 15/Jul/19
Answered by Hope last updated on 15/Jul/19
((((√2) )^2 )/(x+y))+((((√2) )^2 )/(y+z))+((((√2) )^2 )/(z+x))≥((((√2) +(√2) +(√2) )^2 )/(2(x+y+z)))  do≥((9×2)/(2(x+y+z)))  (2/(x+y))+(2/(y+z))+(2/(z+x))≥(9/(x+y+z)) proved
$$\frac{\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} }{{x}+{y}}+\frac{\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} }{{y}+{z}}+\frac{\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} }{{z}+{x}}\geqslant\frac{\left(\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} }{\mathrm{2}\left({x}+{y}+{z}\right)} \\ $$$${do}\geqslant\frac{\mathrm{9}×\mathrm{2}}{\mathrm{2}\left({x}+{y}+{z}\right)} \\ $$$$\frac{\mathrm{2}}{{x}+{y}}+\frac{\mathrm{2}}{{y}+{z}}+\frac{\mathrm{2}}{{z}+{x}}\geqslant\frac{\mathrm{9}}{{x}+{y}+{z}}\:{proved} \\ $$

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