Question Number 151212 by mathdanisur last updated on 19/Aug/21
$$\mathrm{if}\:\:\:\mathrm{a}_{\mathrm{1}} ,\mathrm{a}_{\mathrm{2}} ,...\mathrm{a}_{\boldsymbol{\mathrm{n}}} >\mathrm{1}\:\:\mathrm{then}: \\ $$$$\sqrt{\frac{\left(\mathrm{a}_{\mathrm{1}} -\mathrm{1}\right)\left(\mathrm{a}_{\mathrm{2}} -\mathrm{1}\right)...\left(\mathrm{a}_{\boldsymbol{\mathrm{n}}} -\mathrm{1}\right)}{\left(\mathrm{a}_{\mathrm{1}} +\mathrm{1}\right)\left(\mathrm{a}_{\mathrm{2}} +\mathrm{1}\right)...\left(\mathrm{a}_{\boldsymbol{\mathrm{n}}} +\mathrm{1}\right)}}\:\leqslant\:\frac{\mathrm{a}_{\mathrm{1}} \mathrm{a}_{\mathrm{2}} ...\mathrm{a}_{\boldsymbol{\mathrm{n}}} }{\mathrm{2}^{\boldsymbol{\mathrm{n}}} } \\ $$
Answered by dumitrel last updated on 19/Aug/21
$${x}\geqslant\mathrm{1}\Rightarrow\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}<\frac{{x}}{\mathrm{2}}\Leftrightarrow{x}^{\mathrm{3}} +\left({x}−\mathrm{2}\right)^{\mathrm{2}} >\mathrm{0} \\ $$$$ \\ $$
Commented by mathdanisur last updated on 19/Aug/21
$$\mathrm{Cool}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you} \\ $$