Question Number 13153 by Tinkutara last updated on 15/May/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17
$$\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)\left(\mathrm{1}+{d}\right)= \\ $$$$\left({abcd}+{a}\right)\left({abcd}+{b}\right)\left({abcd}+{c}\right)\left({abcd}+{d}\right)= \\ $$$${abcd}\left({dbc}+\mathrm{1}\right)\left({acd}+\mathrm{1}\right)\left({abd}+\mathrm{1}\right)\left({abc}+\mathrm{1}\right)= \\ $$$$\left({abc}+\mathrm{1}\right)\left({abd}+\mathrm{1}\right)\left({acd}+\mathrm{1}\right)\left({dbc}+\mathrm{1}\right)\geqslant \\ $$$$\geqslant\mathrm{2}\sqrt{{abc}×\mathrm{1}}×\mathrm{2}\sqrt{{abd}×\mathrm{1}}×\mathrm{2}\sqrt{{acd}×\mathrm{1}}×\mathrm{2}\sqrt{{dbc}×\mathrm{1}}= \\ $$$$\geqslant\mathrm{16}{abcd}\sqrt{{abcd}}=\mathrm{16}\:\:.\blacksquare \\ $$
Commented by Tinkutara last updated on 16/May/17
$$\mathrm{Thanks}\:\mathrm{behi}! \\ $$
Commented by RasheedSindhi last updated on 20/May/17
$$\mathcal{V}\:\mathcal{N}{ice}!!! \\ $$
Answered by mrW1 last updated on 16/May/17
$${since}\:\frac{{x}+{y}}{\mathrm{2}}\geqslant\sqrt{{xy}},\:{we}\:{have} \\ $$$$\mathrm{1}+{a}\geqslant\mathrm{2}\sqrt{{a}} \\ $$$$\mathrm{1}+{b}\geqslant\mathrm{2}\sqrt{{b}} \\ $$$$\mathrm{1}+{c}\geqslant\mathrm{2}\sqrt{{c}} \\ $$$$\mathrm{1}+{d}\geqslant\mathrm{2}\sqrt{{d}} \\ $$$$\Rightarrow\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)\left(\mathrm{1}+{d}\right)\geqslant\mathrm{2}^{\mathrm{4}} \sqrt{{abcd}}=\mathrm{16}\sqrt{\mathrm{1}}=\mathrm{16} \\ $$
Commented by Tinkutara last updated on 16/May/17
$$\mathrm{Thanks}\:\mathrm{mrW1}! \\ $$
Commented by RasheedSindhi last updated on 20/May/17
$$\mathcal{N}{ice}\:\&\:\mathcal{S}{mart}!!! \\ $$