Regarding AM-GM inequality

Question

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 $a_1=a_2=\ldots=a_n$.

Answer 1

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 $$\frac{a+b+c+d}{4}=\frac{\frac{a+b}{2}+\frac{c+d}{2}}{2}\geq\frac{2\sqrt{ab}+2\sqrt{cd}}{4}=$$ $$=\frac{\sqrt{ab}+\sqrt{cd}}{2}\geq\frac{2\sqrt[4]{abcd}}{2}=\sqrt[4]{abcd}.$$ 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: $$\frac{a+b+c+\frac{a+b+c}{3}}{4}\geq\sqrt[4]{abc\cdot\frac{a+b+c}{3}}.$$ 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 $n$.

Answer 2

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 $(a_i, a_j) \to (a, a)$?

Answer 3

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 $$ \frac1n\sum_{k=1}^na_k\ge\left(\prod_{k=1}^na_k\right)^{1/n}\tag2 $$

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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.