Why $a^mb^nc^p....$ depends on $\left(\frac{a}{m}\right)^{m}\left(\frac{b}{n}\right)^{n}\left(\frac{c}{p}\right)^{p} \ldots$ for being the greatest?

Question

I am studying Higher Algebra by Hall and Knight and not much explanation is given on any article. So, I had some doubts on this article.

To find the greatest value of $a^mb^nc^p....$ when $a+b+c+.....$ is constant; $m,n,p.....$ being positive integers.

Since $m,n,p.....$ are constants the expression $a^mb^nc^p....$ will be greatest when $\left(\frac{a}{m}\right)^{m}\left(\frac{b}{n}\right)^{n}\left(\frac{c}{p}\right)^{p} \ldots$ is greatest. (For complete article an image is attached)

Now, here I am not able to understand why $a^mb^nc^p....$ depends on $\left(\frac{a}{m}\right)^{m}\left(\frac{b}{n}\right)^{n}\left(\frac{c}{p}\right)^{p} \ldots$ for being the greatest?

Also, can anyone please explain how the author got $m^{n} n^{n} p^{p} \ldots\left(\frac{a+b+c+\ldots}{m+n+p+\ldots}\right)^{m+n+p+\ldots}$ as the greatest value?

Any help would be appreciated.

Answer 1

Because $ m^m n^n \ldots $ is a constant, so you can multiply/divide/add/subtract it and still retain the relative sizing.

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AM-GM gives the equality condition easily. Did they talk about it before?

Answer 2

Hint: Since $m,n,k,\dots$ are constant, $a^mb^nc^k\dots$ is $(a/m)^m(b/n)^n\dots$ multiplied by a (positive) constant. So one is maximum iff the other is. Then AM-GM inequality:

$$\begin{aligned} &\phantom{=} a + b + \cdots \\ &=\underbrace{a/m + a/m + \cdots + a/m}_m + \underbrace{b/n + b/n + \cdots + b/n}_n + \cdots \\ &\ge ... \end{aligned}$$

Answer 3

I was reading the Hall-Knight Higher Algebra book, came across this section a couple of days ago and had the same question. Anyway, answering your questions:

Now, here I am not able to understand why $a^mb^nc^p\cdots$ depends on $\left(\frac{a}{m}\right)^m\left(\frac{b}{n}\right)^n\left(\frac{c}{p}\right)^p\cdots$ for being the greatest?

It doesn't.

Since $m^mn^np^p\cdots$ is a constant value, say $k$, maxing out $\frac{a^mb^nc^p\cdots}{k}$ implies maxing out $a^mb^nc^p\cdots$ and it helps with algebraic gymnastics to arrive at the required solution.

But approaching the solution from this angle does not help much as I experienced.

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Also, can anyone please explain how the author got $m^mn^np^p\cdots\left(\frac{a+b+c+\cdots}{m+n+p+\cdots}\right)^{m+n+p+\cdots}$ as the greatest value?

Fellow victim of Hall-Knight Higher Algebra text book, I shall help thee...

Since the book says $a+b+c+\cdots$ is a constant, we shall denote it by $s$ $$ a+b+c+\cdots=s $$

This can be re-written as, $$ \begin{aligned} s&=m\left(\frac{a}{m}\right)+n\left(\frac{b}{n}\right)+p\left(\frac{c}{p}\right)+\cdots \\ &=\underbrace{\left(\frac{a}{m}+\frac{a}{m}+\cdots\right)}_\text{$m \text{ times}$} + \underbrace{\left(\frac{b}{n}+\frac{b}{n}+\cdots\right)}_\text{$n \text{ times}$} + \underbrace{\left(\frac{c}{p}+\frac{c}{p}+\cdots\right)}_\text{$p \text{ times}$} + \cdots \end{aligned} $$

From the general AM-GM inequality we have, $$ \left(\frac{a_1+a_2+\cdots+a_r}{r}\right)^r \geq a_1a_2\cdots a_r $$ Applying this to the previous equation on $s$, we get $$ \left(\frac{\frac{a}{m}+\frac{a}{m}+\cdots+\frac{b}{n}+\frac{b}{n}+\cdots+\frac{c}{p}+\frac{c}{p}+\cdots}{m+n+p+\cdots}\right)^{m+n+p+\cdots} \geq \left(\frac{a}{m}\right)^m\left(\frac{b}{n}\right)^n\left(\frac{c}{p}\right)^p\cdots $$

Notice that the numerator on the LHS is $s$. Upon simplification we get, $$ m^mn^np^p\cdots\times\left(\frac{a+b+c+\cdots}{m+n+p+\cdots}\right)^{m+n+p+\cdots} \geq a^mb^nc^p\cdots $$

From the above result, it is clear that the max. value of $a^mb^nc^p\cdots$ is $$ m^mn^np^p\cdots\left(\frac{a+b+c+\cdots}{m+n+p+\cdots}\right)^{m+n+p+\cdots} $$ and we're done.

PS: I personally think that starting the solution by maxing out $\left(\frac{a}{m}\right)^m\left(\frac{b}{n}\right)^n\left(\frac{c}{p}\right)^p\cdots$ is a bit confusing.