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Question Number 151212 by mathdanisur last updated on 19/Aug/21
if   a_1 ,a_2 ,...a_n >1  then:  (√(((a_1 -1)(a_2 -1)...(a_n -1))/((a_1 +1)(a_2 +1)...(a_n +1)))) ≤ ((a_1 a_2 ...a_n )/2^n )
$$\mathrm{if}\:\:\:\mathrm{a}_{\mathrm{1}} ,\mathrm{a}_{\mathrm{2}} ,…\mathrm{a}_{\boldsymbol{\mathrm{n}}} >\mathrm{1}\:\:\mathrm{then}: \\ $$$$\sqrt{\frac{\left(\mathrm{a}_{\mathrm{1}} -\mathrm{1}\right)\left(\mathrm{a}_{\mathrm{2}} -\mathrm{1}\right)…\left(\mathrm{a}_{\boldsymbol{\mathrm{n}}} -\mathrm{1}\right)}{\left(\mathrm{a}_{\mathrm{1}} +\mathrm{1}\right)\left(\mathrm{a}_{\mathrm{2}} +\mathrm{1}\right)…\left(\mathrm{a}_{\boldsymbol{\mathrm{n}}} +\mathrm{1}\right)}}\:\leqslant\:\frac{\mathrm{a}_{\mathrm{1}} \mathrm{a}_{\mathrm{2}} …\mathrm{a}_{\boldsymbol{\mathrm{n}}} }{\mathrm{2}^{\boldsymbol{\mathrm{n}}} } \\ $$
Answered by dumitrel last updated on 19/Aug/21
x≥1⇒(√((x−1)/(x+1)))<(x/2)⇔x^3 +(x−2)^2 >0
$${x}\geqslant\mathrm{1}\Rightarrow\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}<\frac{{x}}{\mathrm{2}}\Leftrightarrow{x}^{\mathrm{3}} +\left({x}−\mathrm{2}\right)^{\mathrm{2}} >\mathrm{0} \\ $$$$ \\ $$
Commented by mathdanisur last updated on 19/Aug/21
Cool Ser, thank you
$$\mathrm{Cool}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you} \\ $$

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