# Regarding AM-GM inequality

I have to show that if $$a_1,a_2,\ldots a_n$$ are non-negative real numbers, then $$\frac{a_1+a_2+\ldots+ a_n}{n}\geq (a_1a_2\ldots a_n)^{1/n}$$. Also equality holds if and only if $$a_1=a_2=\ldots=a_n$$.

The inequality can be proved by induction in two steps, viz. First we prove that this is true for all $$n=2^k$$ and for all $$n\in \mathbb{N}$$. But how to show equality part? One way it is clear. But how to show that $$\frac{a_1+a_2+\ldots+ a_n}{n} =(a_1a_2\ldots a_n)^{1/n}$$ implies $$a_1=a_2=\ldots=a_n$$.

## 3 Answers

Because in the base of the induction you used that $$\frac{a+b}{2}\geq\sqrt{ab}$$ for non-negatives $$a$$ and $$b$$.

But it's $$a+b-2\sqrt{ab}\geq0$$ or $$(\sqrt{a}-\sqrt{b})^2\geq0$$ and we see that the equality occurs for $$a=b$$.

Now, we can see that for all $$n=2^k$$, where $$k$$ is a natural number, the equality occurs for $$a_1=a_2=...=a_n.$$

For example, for non-negatives $$a,$$ $$b$$, $$c$$ and $$d$$ by the assumption of the induction we have $$\frac{a+b+c+d}{4}=\frac{\frac{a+b}{2}+\frac{c+d}{2}}{2}\geq\frac{2\sqrt{ab}+2\sqrt{cd}}{4}=$$ $$=\frac{\sqrt{ab}+\sqrt{cd}}{2}\geq\frac{2\sqrt{abcd}}{2}=\sqrt{abcd}.$$ The equality occurs for $$a=b$$, $$c=d$$ and $$ab=cd,$$ which gives $$a=b=c=d.$$

For all natural $$n$$ the reasoning is the same.

For example, for $$n=3$$ we obtain: $$\frac{a+b+c+\frac{a+b+c}{3}}{4}\geq\sqrt{abc\cdot\frac{a+b+c}{3}}.$$ Since for $$n=2^k$$ the equality occurs for $$a_1=a_2=...=a_n,$$

we see that in the last inequality the equality occurs for $$a=b=c=\frac{a+b+c}{3},$$ which gives $$a=b=c$$.

Similarly, for all natural $$n$$.

Suppose that equality holds but $$a_i \ne a_j$$ where $$i \ne j;$$ let $$a = 0.5(a_i+a_j).$$ What do you notice about the transformation $$(a_i, a_j) \to (a, a)$$?

Since $$\log(x)$$ is a concave function, Jensen's Inequality says $$\log\left(\frac1n\sum_{k=1}^na_k\right)\ge\frac1n\sum_{k=1}^n\log(a_k)\tag1$$ which is equivalent to $$\frac1n\sum_{k=1}^na_k\ge\left(\prod_{k=1}^na_k\right)^{1/n}\tag2$$

Suppose that $$\prod_{k=1}^na_k=1\tag3$$ For any direction $$\delta a_k$$ we perturb $$a_k$$, we must have \begin{align} 0 &=\delta\left(\prod_{k=1}^na_k\right)\\ &=\prod_{k=1}^na_k\sum_{k=1}^n\frac{\delta a_k}{a_k}\tag4 \end{align} That same perturbation gives $$\delta\left(\frac1n\sum_{k=1}^na_k^n\right)=\sum_{k=1}^na_k^{n-1}\delta a_k\tag5$$ If $$\frac1{a_k}$$ is not parallel to $$a_k^{n-1}$$, there is a perturbation $$\delta a_k$$ which satisfies $$(4)$$ yet for which $$(5)$$ is non-zero; thus, $$a_k$$ can not be critical. So that the sum is minimal, we need $$a_k^{n-1}=\frac\lambda{a_k}$$ for some $$\lambda$$. Using $$(3)$$, we get $$a_k=1$$. When scaled back for an arbitrary product, we get that all of the $$a_k$$ must be equal.