Regarding the specifity of the A.M.-G.M. inequality in finding maximum and minimum in Number Theory

Question

Today, my math teacher solved a problem which asked to find the maximum value of the expression $x^2y^3$ when $x$ and $y$ are related as $3x+4y=5$.

It was solved using the classic A.M.-G.M. inequality by splitting $3x$ as twice of $3x/2$ and $4y$ as thrice of $4y/3$.

He finally got an inequality as $x^2y^3\le3/16$ and he directly mapped the maximum value of the expression to 3/16 by solely depending only on this inequality. I don't get this point of assigning its maximum value. What if some other inequality says that it is even lesser than 3/16 or restricts its range even further to numbers which are less than 3/16. Can't we give any counterexample inequalities? Or won't any such inequality ever exist? If yes, then why?

I'm a pre-calculus student and I'm still a beginner and am sorry if my question is too dishonorable.

Answer 1

The AM-GM inequality says that for non-negative reals $x_1,x_2,\ldots,x_k$ we always have $$\frac{x_1+x_2+\cdots+x_k}{k}\geq \sqrt[k]{x_1x_2\cdots x_k},$$ with equality if and only if $x_1=x_2=\cdots=x_k$.

Here the result says we have equality if and only if $3x/2=4y/3$. Since this can happen (if $x=2/3$ and $y=3/4$), the upper bound we get is actually the maximum.

Answer 2

If you can prove $x^2y^3 \leqslant 3/16$ [which AM-GM does with the restriction to positive numbers] AND show that equality can be achieved for some value [as it does in this case for $(x_0, y_0) = (\frac23, \frac34)$]; then we can conclude $3/16$ is the maximum.

In general if you can show $f(x, y,...) \leqslant C=f(x_0, y_0, ...)$ for some fixed real $C$ and allowable $(x_0, y_0,...)$, then you have the global maximum as $C$. (Similarly for the minimum, with the sign reversed.).

Note both conditions are required, and together they practically define what a global maximum (or minimum) is.

Answer 3

AM-GM tells us that $x^2y^3 \le \frac{3}{16}$. It also has an equality case, when all of the terms in the AM-GM are equal—in your case, that would be when $\frac{3x}{2} = \frac{4y}{3}$. Combining this with the assumption $3x + 4y = 5$, this gives us only one pair where equality holds: $(x, y) = (\tfrac{2}{3},\tfrac{3}{4})$.

So, if I understand your question correctly, you’re asking why we can outright say that $\frac{3}{16}$ is the maximum. After all, some stronger inequality might be able to prove that $x^2y^3$ is always less than a value less than $\frac{3}{16}$.

The word ‘maximum’ essentially consists of two parts. So, when your teacher is saying that $x^2y^3$ has a maximum value of $\frac{3}{16}$, he/she is saying two things:

1. $x^2y^3 \le \frac{3}{16}$ for all positive reals $x$ and $y$ such that $3x + 4y = 5$.
2. There exists a pair $(x, y)$ that satisfies $3x + 4y = 5$ such that $x^2y^3 = \frac{3}{16}$.

The second condition is equivalent to saying that the bound of $\frac{3}{16}$ can’t be reduced any further.

So, in our case, are both conditions met? By AM-GM, the first condition is met, as you noted. The second condition is also met because we have found values of $x$ and $y$, namely $\frac{2}{3}$ and $\frac{3}{4}$, where equality is met, as $(\frac{2}{3})^2(\frac{3}{4})^3 = \frac{3}{16}$.

I hope this made sense.