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Question-151791




Question Number 151791 by mathdanisur last updated on 23/Aug/21
Answered by ghimisi last updated on 23/Aug/21
x_i ^4 +1≥2x_i ^2 ⇒x_i ^4 −x_i ^2 +1≥x_i ^2 ⇒(√(x_i ^4 −x_i ^2 +1))+2x_i ≥3x_i   Σ_(i=1) ^n ((√(x_i ^4 −x_i ^2 +1))+2x_i )≥Σ_(i=1) ^n 3x_i =3Σx_i ≥3∙n((x_1 x_2 ...x_n ))^(1/n) =3n
$${x}_{{i}} ^{\mathrm{4}} +\mathrm{1}\geqslant\mathrm{2}{x}_{{i}} ^{\mathrm{2}} \Rightarrow{x}_{{i}} ^{\mathrm{4}} −{x}_{{i}} ^{\mathrm{2}} +\mathrm{1}\geqslant{x}_{{i}} ^{\mathrm{2}} \Rightarrow\sqrt{{x}_{{i}} ^{\mathrm{4}} −{x}_{{i}} ^{\mathrm{2}} +\mathrm{1}}+\mathrm{2}{x}_{{i}} \geqslant\mathrm{3}{x}_{{i}} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\sqrt{{x}_{{i}} ^{\mathrm{4}} −{x}_{{i}} ^{\mathrm{2}} +\mathrm{1}}+\mathrm{2}{x}_{{i}} \right)\geqslant\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{3}{x}_{{i}} =\mathrm{3}\Sigma{x}_{{i}} \geqslant\mathrm{3}\centerdot{n}\sqrt[{{n}}]{{x}_{\mathrm{1}} {x}_{\mathrm{2}} …{x}_{{n}} }=\mathrm{3}{n} \\ $$
Commented by mathdanisur last updated on 23/Aug/21
Thank you Ser nice
$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{nice} \\ $$

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