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## Proof: (related to "Inequality Between the Geometric and the Arithmetic Mean")

• By hypothesis, $a_1,\ldots,a_n$ are non-negative positive real numbers.
• If at least one $a_i=0$, then the whole product $a_1\cdots a_n=0$ and the inequation between the geometric mean and their arithmetic mean $\sqrt[n]{a_1\cdots a_n}\le \frac{a_1+\cdots+a_n}n$ is trivial.
• Therefore, we assume $a_1\cdots a_n > 0$ and prove the equivalent inequation ${a_1\cdots a_n}\le \left(\frac{a_1+\cdots+a_n}n\right)^n$ by induction on $n.$
• Base Case:
• For $n=1$ the inequation is true: $a_1 \le \left(\frac{a_1}{1}\right)^1=a_1.$
• Induction Step:
• Assume, the inequation ${a_1\cdots a_n}\le \left(\frac{a_1+\cdots+a_n}n\right)^n$ is true for $n$ numbers.
• For $n+1$ positive numbers $a_1,\ldots,a_n,a_{n+1}$ we can assume without loss of generality that $a_{n+1}\ge a_i$ for $i=1,\ldots,n.$
• It follows from the inequality for weighted arithmetic mean that $\alpha:=\frac{a_1+\cdots+a_n}n\le a_{n+1}$, which means that $a_{n+1}-\alpha\ge 0$ and it follows (since $(n+1)\alpha$ is positive) $$x:=\frac{a_{n+1}-\alpha}{(n+1)\alpha}\ge 0.$$
• Note that $(n+1)\alpha+a_{n+1}-\alpha=a_1+\cdots+a_{n+1}.$ Therefore we get with Bernoulli’s inequality $$\begin{array}{rcl}\left(\frac{a_1+\cdots+a_{n+1}}{(n+1)\alpha}\right)^{n+1}&=&\left(\frac{(n+1)\alpha}{(n+1)\alpha}+\frac{a_{n+1}-\alpha}{(n+1)\alpha}\right)^{n+1}\\&=&(1+x)^{n+1}\\&\ge& 1+(n+1)x\\&=&\frac{a_{n+1}}{\alpha}\end{array}.$$
• This is equivalent to $$\begin{array}{rcl}\left(\frac{a_1+\cdots+a_{n+1}}{(n+1)}\right)^{n+1}&\ge&\alpha^{n+1}\frac{a_{n+1}}{\alpha}=\alpha^n a_{n+1}\end{array}.$$
• Since by the induction assumption $\alpha^n\ge a_1\cdots a_n,$ we get finally $$\begin{array}{rcl}\left(\frac{a_1+\cdots+a_{n+1}}{(n+1)}\right)^{n+1}&\ge&a_1\cdots a_n a_{n+1}\end{array}.$$
q.e.d

| | | | created: 2020-02-02 22:02:04 | modified: 2020-02-02 22:02:55 | by: | references: [586]

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