Previous in Matrices and Determinants Next in Matrices and Determinants
Question Number 88364 by Power last updated on 10/Apr/20
Answered by mind is power last updated on 10/Apr/20
$${by}\:{Cauchy}\:{shwart} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{1}.{x}_{{i}} \leqslant\sqrt{\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}}\mathrm{1}^{\mathrm{2}} .\sqrt{\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{x}_{{i}} ^{\mathrm{2}} },.....\mathrm{1} \\ $$$$\Sigma{x}_{{i}} =\mathrm{1} \\ $$$$\Sigma{x}_{{i}} ^{\mathrm{2}} =\frac{\mathrm{1}}{{n}} \\ $$$$\mathrm{1}\Leftrightarrow\mathrm{1}\leqslant\sqrt{{n}}.\frac{\mathrm{1}}{\sqrt{{n}}}=\mathrm{1}\:{equality}\Rightarrow{by}\:{cauchy}\:{shwartz} \\ $$$$\left({x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,....,{x}_{{n}} \right)=\left({a},{a},....,{a}\right) \\ $$$$\Rightarrow{na}=\mathrm{1}\Rightarrow{a}=\frac{\mathrm{1}}{{n}},\forall{i}\in\left\{\mathrm{1},....{n}\right\}\:{x}_{{i}} =\frac{\mathrm{1}}{{n}} \\ $$