Question Number 58462 by naka3546 last updated on 23/Apr/19
$${a},\:{b},\:{c},\:{d}\:\:\in\:\:\mathbb{R}^{+} \\ $$$${a}\:+\:{b}\:+\:{c}\:+\:{d}\:\:=\:\:\mathrm{1} \\ $$$${Prove}\:\:{that}\:\:: \\ $$$${abc}\:+\:{bcd}\:+\:{cda}\:+\:{dab}\:\:\leqslant\:\:\frac{\mathrm{1}}{\mathrm{27}}\:\:+\:\:\frac{\mathrm{176}}{\mathrm{27}}\:{abcd} \\ $$
Answered by tanmay last updated on 24/Apr/19
$$\frac{{a}+{b}+{c}+{d}}{\mathrm{4}}\geqslant\left({abcd}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${abcd}\leqslant\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{4}} \rightarrow{abcd}\leqslant\frac{\mathrm{1}}{\mathrm{256}} \\ $$$${considering}\:\:{abcd}=\frac{\mathrm{1}}{\mathrm{256}} \\ $$$$\left({abc}+{bcd}+{cda}+{dab}\right)×\frac{\mathrm{1}}{\mathrm{4}}\geqslant\left({a}^{\mathrm{3}} {b}^{\mathrm{3}} {c}^{\mathrm{3}} {d}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\left({abc}+{bcd}+{cda}+{dab}\right)\geqslant\mathrm{4}\left({abcd}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$\left({abc}+{bcd}+{cda}+{dab}\right)\geqslant\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{256}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$ \\ $$$$\left({abc}+{bcd}+{cda}+{dab}\right)\geqslant\frac{\mathrm{1}}{\mathrm{16}} \\ $$$${abc}+{bcd}+{cda}+{dab}=\frac{\mathrm{1}}{\mathrm{16}} \\ $$$${RHS} \\ $$$$\frac{\mathrm{1}}{\mathrm{27}}+\frac{\mathrm{176}}{\mathrm{27}}{abcd} \\ $$$$\frac{\mathrm{1}}{\mathrm{27}}+\frac{\mathrm{176}×\mathrm{1}}{\mathrm{27}×\mathrm{256}} \\ $$$$\frac{\mathrm{1}}{\mathrm{27}}\left(\mathrm{1}+\frac{\mathrm{11}}{\mathrm{16}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}} \\ $$$${LHS}={RHS} \\ $$$${i}\:{have}\:{solved}\:{considering}\:\:={sign}\:{in}\geqslant\:{sign} \\ $$$$ \\ $$