I'm reading Byrne's "A first course in optimization". After the presentation and proof of the AMGM inequality, the author gives two examples:
Here's the first:
I guess I understand what was done, it seems he writes in a convenient form with the AMGM inequality:
$$\left(\frac{3\cdot12}{x} \cdot\frac{3\cdot18}{y} \cdot 3\cdot xy\right)^{\frac{1}{3}}\leq \frac{1 }{3} \left( \frac{3\cdot12}{x} +\frac{3\cdot18}{y} + 3\cdot xy\right)$$
Which in turn, yields:
$$\left(3^3\cdot 12 \cdot 18\right)^{\frac{1}{3}}=\left(3^3 2^3 3^3\right)^{\frac{1}{3}}=18\leq \frac{12}{x} +\frac{18}{y} + xy$$
From here, I know that for all possible choices of $x,y$, then $18$ is the smaller value, but I'm not sure on how to find the actual $x,y$. Given that he mentions that the smaller value for $\frac{1}{3} f(x,y)$ is $6$, then I guess he did the following:
$$\frac{1}{3} f(x,y)+\frac{1}{3} f(x,y)+\frac{1}{3} f(x,y) \leq 6+6+6=18$$
$$\overbrace{\left(\frac{4}{x} +\frac{6}{y} + \frac{xy}{3}\right)}^{= \;6}+\overbrace{\left(\frac{4}{x} +\frac{6}{y} + \frac{xy}{3}\right)}^{=\;6}+\overbrace{\left(\frac{4}{x} +\frac{6}{y} + \frac{xy}{3}\right)}^{=\;6}\leq 6+6+6=18$$
$$\overbrace{\left(\frac{4}{x} +\frac{4}{x} + \frac{4}{x}\right)}^{= \;6}+\overbrace{\left(\frac{6}{y} +\frac{6}{y} + \frac{6}{y}\right)}^{= \;6}+\overbrace{\left(\frac{xy}{3} +\frac{xy}{3} + \frac{xy}{3}\right)}^{= \;6}=\frac{12}{x} +\frac{18}{y} + xy\leq 18$$
And it seems that solving each of them individually yields the desired result, I am confused at why reorganizing $3$ sums whose total is $6$ in this way (supposing it is what actually was done) yields exactly the the minimal solution.
My doubt in the following example may stem from the same doubt I had before:
Here, he talks about a "constant sum" which is perhaps something very similar to what was done before, but I don't understand what is this "constant sum" nor why it is needed, I see that the second centered equation is the same as the first but I don't understand the workings of this "constant sum". I have tried to rewrite the equation in several other ways (on paper) to check if I could uncover something but I couldn't. Could you clarify?
In both cases it was used: $$a+b+c\ge 3\sqrt[3]{abc},$$ the equality occurs when $a=b=c$.
Example 1: $$\frac{12}{x}+\frac{18}{y}+xy\ge 3\sqrt[3]{\frac{12}{x}\cdot \frac{18}{y}\cdot xy}=18,$$ $$\frac{12}{x}=\frac{18}{y}=xy \Rightarrow x=2, y=3$$ Example 2: $$\frac{1}{12}\cdot (3x)(4y)(72-3x-4y)\le \frac{1}{12}\left(\frac{3x+4y+72-3x-4y}{3}\right)^3=1152, $$ $$3x=4y=72-3x-4y \Rightarrow x=8, y=6.$$