Question Number 194691 by CrispyXYZ last updated on 13/Jul/23
$${a},\:{b},\:{c}\geqslant\mathrm{0},\:{a}+{b}+{c}=\mathrm{2}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{3}{a}+\mathrm{8}{ab}+\mathrm{16}{abc}\leqslant\mathrm{12}. \\ $$
a-b-c-0-a-b-c-2-Prove-that-3a-8ab-16abc-12-