Question Number 7782 by Chantria last updated on 15/Sep/16
$$\:{Given}\:{a},{b},{c}\:\in{N}\:;\:{prove}\:{that} \\ $$$$\:\frac{\mathrm{1}+{a}}{\mathrm{1}+\mathrm{2}{a}}\:+\:\frac{\mathrm{1}+{b}}{\mathrm{1}+\mathrm{2}{b}}\:+\:\frac{\mathrm{1}+{c}}{\mathrm{1}+\mathrm{2}{c}}\:\leqslant\:\mathrm{2} \\ $$
Commented by sou1618 last updated on 15/Sep/16
$${Let}\:{f}\left({n}\right)=\frac{\mathrm{1}+{n}}{\mathrm{1}+\mathrm{2}{n}}\:\:\left({n}\geqslant\mathrm{1}\right) \\ $$$$ \\ $$$${f}\:'\left({n}\right)=\frac{\mathrm{1}×\left(\mathrm{1}+\mathrm{2}{n}\right)−\left(\mathrm{1}+{n}\right)×\mathrm{2}}{\left(\mathrm{1}+\mathrm{2}{n}\right)^{\mathrm{2}} } \\ $$$${f}\:'\left({n}\right)=\frac{−\mathrm{1}}{\left(\mathrm{1}+\mathrm{2}{n}\right)^{\mathrm{2}} }<\mathrm{0} \\ $$$${so} \\ $$$${f}\left({n}\right)\leqslant{f}\left(\mathrm{1}\right)\:\:\:\:\:\:\left({n}\geqslant\mathrm{1}\right) \\ $$$${f}\left(\mathrm{1}\right)=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$ \\ $$$$\frac{\mathrm{1}+{a}}{\mathrm{1}+\mathrm{2}{a}}+\frac{\mathrm{1}+{b}}{\mathrm{1}+\mathrm{2}{b}}+\frac{\mathrm{1}+{c}}{\mathrm{1}+\mathrm{2}{c}}\leqslant\mathrm{3}{f}\left({n}\right)\leqslant\mathrm{2} \\ $$$$ \\ $$