Statement:

  Let \(\pmb{a_{1},a_{2},…,a_{n}} \) be \(\pmb{n} \) positive real numbers then \(\pmb{ \frac{a_{1}+a_{2}+…+a_{n}}{n}\geq \sqrt[n]{a_{1}a_{2}…a_{n}} }\). The equality occurs when \(\pmb{a_{1}=a_{2}=…=a_{n}} \).

  Proof:
  Let \(\pmb{a_{1},a_{2},…,a_{n}} \) be \(\pmb{n} \) positive real numbers.

• Case-1: Let \(\pmb{n=2^{k} }\) where \(\pmb{k }\) is positive integers.
  For any two positive real numbers \(\pmb{p,q }\)
  \begin{align} &\pmb{pq=\left(\frac{p+q}{2} \right)^{2}-\left(\frac{p-q}{2} \right)^{2}}\nonumber\\ \implies & \pmb{pq \leq \left(\frac{p+q}{2} \right)^{2}} \end{align} Since \(\pmb{ \left(p-q\right)^{2}\geq 0 } \). The equalily occurs if \(\pmb{ p=q } \).
  Putting \( \pmb{ p=a_{1},~q=a_{2}} \) in (1), we have \begin{align} &\pmb{a_{1}a_{2} \leq \left(\frac{a_{1}+a_{2}}{2} \right)^{2}} \\ \implies & \pmb{ \frac{a_{1}+a_{2}}{2}\geq \sqrt[2]{a_{1}a_{2}} } \end{align} The equality occurs when \(\pmb{a_{1}=a_{2}} \).
  Therefore the statement is true for \(\pmb{n=2 }\).
  Putting \( \pmb{ p=a_{3},~q=a_{4}} \) in (1), we have \begin{align} \pmb{a_{3}a_{4} \leq \left(\frac{a_{3}+a_{4}}{2} \right)^{2}} \end{align} Multiplying (2) and (3), we get \begin{align} &\pmb{a_{1}a_{2}a_{3}a_{4} \leq \left(\frac{a_{1}+a_{2}}{2} \right)^{2}\left(\frac{a_{3}+a_{4}}{2} \right)^{2}} \nonumber\\ \implies & \pmb{a_{1}a_{2}a_{3}a_{4} \leq \left[ \left(\frac{a_{1}+a_{2}}{2} \right)\left(\frac{a_{3}+a_{4}}{2} \right)\right]^{2}} \end{align} Putting \( \pmb{ p=\frac{a_{1}+a_{2}}{2} ,~q=\frac{a_{3}+a_{4}}{2}} \) in (1), we get \begin{align} &\pmb{\left(\frac{a_{1}+a_{2}}{2} \right)\left(\frac{a_{3}+a_{4}}{2} \right) \leq \left[\frac{\left(\frac{a_{1}+a_{2}}{2} \right)+\left(\frac{a_{3}+a_{4}}{2} \right)}{2} \right]^{2}} \nonumber\\ \implies & \pmb{\left[\left(\frac{a_{1}+a_{2}}{2} \right)\left(\frac{a_{3}+a_{4}}{2} \right)\right]^{2} \leq \left[\frac{a_{1}+a_{2}+a_{3}+a_{4} }{4} \right]^{4}} \end{align} From (4) and (5), we have \begin{align} & \pmb{a_{1}a_{2}a_{3}a_{4} \leq \left[\frac{a_{1}+a_{2}+a_{3}+a_{4} }{4} \right]^{4} }\nonumber\\ \implies & \pmb{ \frac{a_{1}+a_{2}+a_{3}+a_{4} }{4}\geq \sqrt[4]{a_{1}a_{2}a_{3}a_{4}} } \end{align} The equality occurs when \(\pmb{a_{1}=a_{2}=a_{3}=a_{4}} \).
  Therefore the statement is true for \(\pmb{n=2^{2}=4 }\).
  Continuing this process, after \(\pmb{ k }\) steps, we have for \(\pmb{n=2^{k} }\) \begin{align} \pmb{ \frac{a_{1}+a_{2}+…+a_{n}}{n}\geq \sqrt[n]{a_{1}a_{2}…a_{n}} } \end{align} The equality occurs when \(\pmb{a_{1}=a_{2}=…=a_{n}} \).
  

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• Case-2: Let \(\pmb{n\ne 2^{k} }\).
  Then \(\pmb{2^{k-1} \lt n \lt 2^{k} }\). Let \(\pmb{ 2^{k}=n+t }\) for some positive integers \(\pmb{t} \).
  Let \begin{align} \pmb{ m=\frac{a_{1}+a_{2}+…+a_{n}}{n} } \end{align} Consider the \(\pmb{n+t} \) positive real numbers \(\pmb{a_{1},a_{2},…,a_{n}} \) and \(\pmb{m,m,…} \) \(\pmb{t} \)-times
  we have for \(\pmb{n+t 2^{k}} \) positive real numbers,
  \begin{align} \pmb{ \frac{a_{1}+a_{2}+…+a_{n}+tm}{n+t}\geq \sqrt[n+t]{a_{1}a_{2}…a_{n}.m^{t}} } \end{align} Substituting (9) in (10), we get \begin{align*} &\pmb{ \frac{nm+tm}{n+t}\geq \sqrt[n+t]{a_{1}a_{2}…a_{n}.m^{t}}}\\ \implies & \pmb{ m\geq \sqrt[n+t]{a_{1}a_{2}…a_{n}.m^{t}} }\\ \implies & \pmb{ m^{n+t}\geq a_{1}a_{2}…a_{n}.m^{t} }\\ \implies & \pmb{ m^{n}m^{t}\geq a_{1}a_{2}…a_{n}.m^{t} }\\ \implies & \pmb{ m^{n}\geq a_{1}a_{2}…a_{n} }\\ \implies & \pmb{ m\geq \sqrt[n]{a_{1}a_{2}…a_{n}} }\\ \implies & \pmb{ \frac{a_{1}+a_{2}+…+a_{n}}{n}\geq \sqrt[n]{a_{1}a_{2}…a_{n}} } \end{align*} The equality occurs when \(\pmb{a_{1}=a_{2}=…=a_{n}} \).
  

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  Therefore \( \pmb{A.M.\geq G.M.} \)