Question Number 145626 by mathdanisur last updated on 06/Jul/21
Commented by mr W last updated on 06/Jul/21
$$\mid{a}+{b}\mid\leqslant\mid{a}\mid+\mid{b}\mid \\ $$$$\frac{\mid{a}\mid+\mid{b}\mid}{\mid{a}+{b}\mid}\geqslant\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mid{a}\mid+\mid{b}\mid}{\mid{a}+{b}\mid}\geqslant\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}=\frac{\mathrm{3}}{\mathrm{2}}={min} \\ $$
Commented by mathdanisur last updated on 06/Jul/21
$${cool}\:{Ser},\:{thaks} \\ $$