1 problem found
Suppose that \(a_{i}>0\) for all \(i>0\). Show that \[ a_{1}a_{2}\leqslant\left(\frac{a_{1}+a_{2}}{2}\right)^{2}. \] Prove by induction that for all positive integers \(m\) \[ a_{1}\cdots a_{2^{m}}\leqslant\left(\frac{a_{1}+\cdots+a_{2^{m}}}{2^{m}}\right)^{2^{m}}.\tag{*} \] If \(n<2^{m}\), put \(b_{1}=a_{2},\) \(b_{2}=a_{2},\cdots,b_{n}=a_{n}\) and \(b_{n+1}=\cdots=b_{2^{m}}=A\), where \[ A=\frac{a_{1}+\cdots+a_{n}}{n}. \] By applying \((*)\) to the \(b_{i},\) show that \[ a_{1}\cdots a_{n}A^{(2^{m}-n)}\leqslant A^{2^{m}} \] (notice that \(b_{1}+\cdots+b_{n}=nA).\) Deduce the (arithmetic mean)/(geometric mean) inequality \[ \left(a_{1}\cdots a_{n}\right)^{1/n}\leqslant\frac{a_{1}+\cdots+a_{n}}{n}. \]
Solution: \begin{align*} && 0 &\leqslant (a_1 - a_2)^2 \\ &&&= a_1^2 -2a_1a_2 + a_2^2 \\ &&&= (a_1+a_2)^2 -4a_1a_2 \\ \Leftrightarrow && a_1a_2 &\leqslant \left ( \frac{a_1+a_2}2 \right)^2 \end{align*} Claim: \((*)\) is true Proof: (By induction) We have already proven the base case. Suppose it is true for some \(m\), then consider \(m+1\) \begin{align*} && a_1 \cdots a_{2^m} &\leqslant \left ( \frac{a_1 + \cdots + a_{2^m}}{2^m} \right)^{2^m} \tag{by (*)} \\ && a_{2^m+1} \cdots a_{2^{m+1}} &\leqslant \left ( \frac{a_{2^m+1} + \cdots + a_{2^{m+1}}}{2^m} \right)^{2^m} \tag{by (*)} \\ \Rightarrow && (a_1 \cdots a_{2^m})^{1/2^m} &\leqslant \left ( \frac{a_1 + \cdots + a_{2^m}}{2^m} \right) \\ && (a_{2^m+1} \cdots a_{2^{m+1}})^{1/2^m} &\leqslant \left ( \frac{a_{2^m+1} + \cdots + a_{2^{m+1}}}{2^m} \right) \\ \Rightarrow && (a_1 \cdots a_{2^m})^{1/2^m} \cdot (a_{2^m+1} \cdots a_{2^{m+1}})^{1/2^m} &\leqslant \left ( \frac{ (a_1 \cdots a_{2^m})^{1/2^m} +(a_{2^m+1} \cdots a_{2^{m+1}})^{1/2^m} }{2} \right )^2 \\ &&&\leqslant \left ( \frac{ \frac{a_1 + \cdots + a_{2^m}}{2^m}+\frac{a_{2^m+1} + \cdots + a_{2^{m+1}}}{2^m} }{2} \right )^2 \\ &&&\leqslant \left ( \frac{ a_1 + \cdots + a_{2^m}+a_{2^m+1} + \cdots + a_{2^{m+1}} }{2^{m+1}} \right )^2 \\ \Rightarrow && a_1 \cdots a_{2^{m+1}} &\leqslant \left ( \frac{a_1 + \cdots + a_{2^{m+1}}}{2^{m+1}} \right)^{2^{m+1}} \end{align*} Which is precisely \((*)\) for \(m+1\). Therefore our statement is true by induction. Suppose \(n < 2^m\) and \(b_1 = a_1, b_2 = a_2, \cdots b_n = a_n\) and \(b_{n+1} = \cdots = b_{2^m} = A\) where \(A = \frac{a_1 + \cdots + a_n}{n}\) then \begin{align*} && b_1 \cdots b_n \cdot b_{n+1} \cdots b_{2^m} &\leq \left ( \frac{b_1 + \cdots + b_n + b_{n+1} + \cdots + b_{2^m}}{2^{m}} \right)^{2^m} \\ \Leftrightarrow && a_1 \cdots a_n \cdot A^{2^m-n} &\leq \left ( \frac{a_1 + \cdots + a_n + (2^m-n)A}{2^m} \right)^{2^m} \\ &&&= \left ( \frac{nA + (2^m - n)A}{2^m} \right)^{2^m} \\ &&&= A^{2^m} \\ \Rightarrow && a_1 \cdots a_n &\leq A^n \\ \Rightarrow && (a_1 \cdots a_n)^{1/n} &\leq A = \frac{a_1 + \cdots + a_n}{n} \end{align*}