91Ó°ÊÓ

Given:$AMof{a}_1,a_2,......,a_{n}is\frac{a_1+a_2+....+a_n}{n}.$

And,$GMof{a}_1,a_2,....,a_{n}is\sqrt[n]{a_1{a}_2...a_n}.$

$$\begin{gathered} Letf(x_1,x_2,...x_n)=x_1{x}_2...x_n. \\ And{x}_1+x_2+....+x_n=1. \\ Thenletg(x_1,x_2,...x_n)=x_1+x_2+....+x_n-1. \\ ∇f(x_1,x_2,...x_n)=\left(\frac{∂f}{∂x_1}\right)i_1+\left(\frac{∂f}{∂x_2}\right)i_2+...+\left(\frac{∂f}{∂x_n}\right)i_n. \\ =\left(x_2...x_n\right)i_1+\left(x_1{x}_3...x_n\right)i_2+.....+\left(x_1...x_{n-1}\right)i_n. \\ ∇g(x_1,x_2,...x_n)=\left(\frac{∂g}{∂x_1}\right)i_1+\left(\frac{∂g}{∂x_2}\right)i_2+...+\left(\frac{∂g}{∂x_n}\right)i_n. \\ =i_1+i_2+....+i_n. \\ Now,bymethodofLagrangemultiplierweget, \\ ∇f=\mathrm{λ}∇g. \\ \left(x_2...x_n\right)i_1+\left(x_1{x}_3...x_n\right)i_2+.....+\left(x_1...x_{n-1}\right)i_n=\mathrm{λ}\left(i_1+i_2+....+i_n\right). \\ Comparingcoefficients, \\ {x}_2...x_n=x_1{x}_3...x_n=.....=x_1...x_{n-1} \\ Thismeans,x_i=x_{j}foralli,j. \end{gathered}$$

$$\begin{gathered} Put{x}_i=x_{j}in{x}_1+x_2+....+x_n=1weget, \\ {x}_i=\frac{1}{n},foralli, \\ Therefore,x=\left(\frac{1}{n},\frac{1}{n},....,\frac{1}{n}\right)isthepointwheremaximumoccur. \\ Maximumvaluewillbe: \\ f(x_1,x_2,...x_n)=\frac{1}{n}.\frac{1}{n}.......\frac{1}{n}=\frac{1}{n^n}. \\ Let{x}_i=\frac{a_i}{A},whereA=a_1+a_2+...+a_n \\ f(x_1,x_2,...x_n)â\copyright ½\frac{1}{n^n}, \\ {x}_1{x}_2...x_nâ\copyright ½\frac{1}{n^n}, \\ \frac{a_1}{A}.\frac{a_2}{A}......\frac{a_n}{A}â\copyright ½\frac{1}{n^n}, \\ {a}_1{a}_2...a_nâ\copyright ½\frac{A^n}{n^n}={\left(\frac{A}{n}\right)}^n={\left(\frac{a_1+a_2+...+a_n}{n}\right)}^n \\ \sqrt[n]{a_1{a}_2...a_n}â\copyright ½\left(\frac{a_1+a_2+...+a_n}{n}\right) \\ Hence,Proved. \end{gathered}$$