Problem source

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 source

\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*}