Inequality Proof in Real Numbers: If $p+q=1$ then $pq\le\frac14$

Question

Let $p,q \in \mathbb R$, and $p+q=1$. Prove that $$pq \le \frac{1}{4}.$$

The first thing I did was define 3 possibilities that we can have from $p+q=1$:

1º $p$ or $q$ is negative. Example:

$p=-10$, $q=11$

$-10+11=1$

$(-10)(11)=-110$

So if $p$ or $q$ are negative the inequality $pq$ $\le$ $\frac14$ holds.

2º $p$ or $q$ is $0$, that makes the inequality $pq$ $\le$ $\frac14$ be $0$ $\le$ $\frac14$, the inequality holds.

3º The final possibility I saw was $0$ $\lt$ $p,q$ $\lt 1$, here i just see that the max value from $p,q$ can be $p,q = 1/2$ making $\frac12 + \frac12 = 1$ and $(\frac12) (\frac12) \le \frac14$, but i do not see the elegant way of proving that with other values the inequality holds.

I apologize in advance if I made any mistake or missed something.

Answer 1

Hint: Let $q = 1-p$. Then $pq = p(1-p) = p - p^2$. This is a quadratic - do you know how to find the maximum?

Answer 2

Note that $pq\leqslant\left(\frac{p+q}2\right)^2$, because$$\left(\frac{p+q}2\right)^2-pq=\left(\frac{p-q}2\right)^2.$$Since $\frac{p+q}2=\frac12$…

Answer 3

Observe that $$4pq = (p+q)^2-(p-q)^2=1-(p-q)^2 \le 1.$$

Answer 4

Set up a quadratic equation $x^2+bx+c=0$ having $p, q$ as the roots. We have: $b=-(p+q)=-1$ and $c=pq$ (Vieta formulae). Thus, the discriminant must be non-negative: $(-1)^2-4pq\ge 0$, i.e. $pq\le\frac{1}{4}$.

Answer 5

AM-GM: $ x, y > 0 $
$ \frac{x+y}{2} \geq \sqrt{xy} \Leftrightarrow \frac{(x+y)^2}{4} \geq \sqrt{xy} \Leftrightarrow xy \leq \frac{1}{4}$
The rest is obvious if two numbers are negative, then their sum is negative. If they are of different signs, then the product is negative.