Question Number 125868 by I want to learn more last updated on 14/Dec/20
Answered by mindispower last updated on 14/Dec/20
$$\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{2}{y}}\geqslant\frac{\mathrm{2}}{{x}+{y}}.\Leftrightarrow\left({x}+{y}\right)^{\mathrm{2}} \geqslant\mathrm{4}{xy},{x},{y}>\mathrm{0} \\ $$$$\Leftrightarrow\left({x}−{y}\right)^{\mathrm{2}} \geqslant\mathrm{0}\:\:{true} \\ $$$$\Rightarrow\forall\left({x},{y}\right)\in\mathbb{R}_{+} ^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{2}{y}}\geqslant\frac{\mathrm{2}}{{x}+{y}} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{\mathrm{2}{a}}+\frac{\mathrm{1}}{\mathrm{2}{b}}+\frac{\mathrm{1}}{\mathrm{2}{a}}+\frac{\mathrm{1}}{\mathrm{2}{c}}+\frac{\mathrm{1}}{\mathrm{2}{b}}+\frac{\mathrm{1}}{\mathrm{2}{c}}...{E} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{a}}+\frac{\mathrm{1}}{\mathrm{2}{b}}\geqslant\frac{\mathrm{2}}{{a}+{b}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{a}}+\frac{\mathrm{1}}{\mathrm{2}{c}}\geqslant\frac{\mathrm{2}}{{a}+{c}},\frac{\mathrm{1}}{\mathrm{2}{c}}+\frac{\mathrm{1}}{\mathrm{2}{b}}\geqslant\frac{\mathrm{2}}{{c}+{b}} \\ $$$${E}\Rightarrow\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\geqslant\frac{\mathrm{2}}{{a}+{b}}+\frac{\mathrm{2}}{{a}+{c}}+\frac{\mathrm{2}}{{b}+{c}} \\ $$$$ \\ $$$$ \\ $$
Commented by I want to learn more last updated on 15/Dec/20
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$