What is the maximum product from two fractions who sum to 1?

Question

I'm working on a problem where I need to prove that a posterior variance of binomial data with a uniform prior is less than 1/12 (the variance of the prior). Currently I have $(\frac{y+1}{n+2})$ $(\frac{n-y+1}{n+2})$ $(\frac{1}{n+3})$ For the variance. Since the first two fractions sum to 1, I know conceptually that the maximum product is $\frac14$. However, how can we prove that the maximum product of two fractions who sum to 1 is $\frac14$?

I know from there how to prove the inequality, but I just don't know how to prove the maximum product.

Answer 1

We have two real numbers $0 < a, b$ such that $a + b = 1$, so $b = 1-a$. We want to maximize $a\cdot b$.

Substitution gives $a \cdot b = a(1-a) = a - a^2$. What is the maximum possible value of that expression?

Answer 2

We have two positive numbers $x$ and $y$ such that $x + y = 1$. Also, for any real number, its square is positive, meaning that $$ 0 \leq (\sqrt x - \sqrt y)^2 $$ (I realise this looks like it comes out of nowhere, but as we see below, it's exactly what we need, so bear with me.) We continue working on that inequality to get $$ 0\leq x -2\sqrt{xy} + y = x+y-2\sqrt{xy}\\ 2\sqrt{xy} \leq x + y= 1\\ \sqrt{xy}\leq \frac{1}{2}\\ xy\leq \frac14 $$ so we cannot do better than $0.25$. Is that value possible to reach? Well, note that the first line is an equality (instead of an inequality) when $\sqrt x - \sqrt y = 0$, or in other words, $x = y$. If we have $x = y$ and $x+y = 1$, then we must have $x = y = 0.5$, which does actually give $xy = 0.25$, so we are done.