To find the greatest value of a product whose sum is constant

Question

For a given sum of two numbers;

$$a+b=S$$ To maximize,$$P=ab$$ is attained when the two factors are the same (so we've been taught), i.e.,

$$\left(\frac{a+b}{2}\right)^2{\ge}ab$$ Also, generally, for $n$ numbers, $$\left(\frac{a+b+c+...n}{n}\right)^n {\ge} abc...n$$

Question:

$1.$ I understand the proof that if $a=b$, the product, $\left(\frac{a+b}{2}\right)^2$ is greater than or equal to $ab$, but how is it the greatest product?

$2.$ For the general case, I understand this is also the case for AM-GM inequality but I don't understand it is the maximum product when the sum is constant.

$3.$ How is it the idea of equality extended to maximising the the product $a^mb^nc^p...k^t$ with given sum $a+b+...+k=Z$

Generally, I don't understand how it is the maximum for both the cases.

Answer 1

1,2.

I think you mean that the variables are positives.

If $a+b+c+...=Z$ so by AM-GM $abc...\leq\left(\frac{Z}{n}\right)^n$.

The equality occurs for $a=b=c=...=\frac{Z}{n}$, which says which you wish.

3. By AM-GM we obtain: $$a^mb^nc^p...=m^m\left(\frac{a}{m}\right)^mn^n\left(\frac{b}{n}\right)^n...\leq$$ $$\leq m^mn^n\left(\frac{m\cdot\frac{a}{m}+n\cdot\frac{b}{n}+...}{m+n+...}\right)^{m+n+...}= m^mn^n...\left(\frac{Z}{N}\right)^{N},$$ where $N=m+n+...$.