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Question Number 194691 by CrispyXYZ last updated on 13/Jul/23

a, b, c≥0, a+b+c=2. Prove that  3a+8ab+16abc≤12.

$${a},\:{b},\:{c}\geqslant\mathrm{0},\:{a}+{b}+{c}=\mathrm{2}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{3}{a}+\mathrm{8}{ab}+\mathrm{16}{abc}\leqslant\mathrm{12}. \\ $$

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