Quadratic function with roots in $[0,1]$. Prove that $f(0) \geq \frac49$ or $f(1) \geq \frac49$

Question

Let $a,b$ in $[0,1]$ be such that the polynomial $f(x) = (x-a)(x-b)$ satisfies $f(\tfrac12) \geq \frac1{36}$.

I have found a quite complicated proof of the following inequality using calculus: $$f(0) \geq \frac49 \quad\text{or}\quad f(1) \geq \frac49.$$

Can you find a simple argument? (with geometric flavour if possible)

Answer 1

I'm gonna prove this algebraically. First some observations:

$$f\left(\frac{1}{2}\right) \geq \frac{1}{36}\Leftrightarrow 2ab-(a+b)+\frac{4}{9} \geq 0 \ \ \ \ \ \ \ (1)$$

$$f(0) \geq \frac{4}{9} \Leftrightarrow ab \geq \frac{4}{9}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$

$$f(1) \geq \frac{4}{9} \Leftrightarrow ab-a-b + \frac{5}{9} \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$

Assume for the sake of contradiction that both $(2)$ and $(3)$ are false, that means:

$$ab \leq \frac{4}{9}\text{ and } ab+\frac{5}{9} \leq a+b$$

Notice that from $(1)$ and AM-GM, we have:

$$0 \leq 2ab-(a+b)+\frac{4}{9} \leq 2ab-2\sqrt{ab}+\frac{4}{9}=2\left(\sqrt{ab}-\frac{1}{3}\right)\left(\sqrt{ab}-\frac{2}{3}\right)$$

However, since $ab \leq \dfrac{4}{9}$, we have $\sqrt{ab} \leq \dfrac{2}{3}$. This, together with the above inequality, implies that $\sqrt{ab} \leq \dfrac{1}{3}$. But combining the negation of $(3)$ and $(1)$, we get:

$$ab+\frac{5}{9}\leq a+b \leq 2ab+\frac{4}{9}\Rightarrow ab \geq \frac{1}{9}\Rightarrow \sqrt{ab} \geq \frac{1}{3}$$

a contradiction. Therefore, at least one of $(2)$ or $(3)$ is true.

Answer 2

Note,

$$f(\frac12) =(\frac12-a)(\frac12-b)\ge \frac{1}{36}\implies 2ab-(a+b)+\frac{4}{9} \geq 0$$

Use $a+b\ge 2\sqrt{ab}$ to factorize above inequality as

$$(\sqrt{ab}-\frac{1}{3})(\sqrt{ab}-\frac{2}{3})\ge 0$$

which leads to either $\sqrt{ab}\le\frac{1}{3}$ or $\sqrt{ab}\ge\frac{2}{3}$. Examine the two cases below.

Case 1) $\sqrt{ab}\ge\frac{2}{3}$. Evalute $f(0)$,

$$f(0) = (0-a)(0-b) = ab \ge \frac49$$

Case 2) $\sqrt{ab}\le\frac{1}{3}$ leads to $-ab \ge -\frac19$. Take $f'(x) = 2x-(a+b)$ and evalute $f(1)$ in two ways,

$$f(1) = f(\frac12) + \int_{1/2}^1 f'(x)dx=f(\frac12)+ \frac34 -\frac12(a+b)$$

$$f(1) = 1-(a+b)+ab$$

Eliminate $a+b$ and apply $f(\frac12)\ge \frac1{36}$, $-ab \ge -\frac19$ to obtain $$f(1)=2f(\frac12)+\frac12-ab\ge \frac2{36}+\frac12 - \frac19 = \frac49 \implies f(1) \ge \frac49$$

Answer 3

From $f(\frac12)\ge\dfrac{1}{36}\gt0$ it is clear that the roots $a$ and $b$ of $f$ are both either, to the left or to the right of $x=\dfrac12$. When $0\le a\le b\le\dfrac12$, the value $f(0)$ is less than $f(1)$ which is easy to see and, by symmetry, it is enough to study this case (in the other case, $\dfrac12\le a\le b\le1$ we obviously have $f(0)\gt f(1)$ which it is a very elementary observation).

So with the data $$\begin{cases}f(x)=x^2-(a+b)x+ab\\0\le a\le b\le\dfrac12\\f\left(\frac12\right)\ge\dfrac{1}{36}\end{cases}$$ we prove that $f(1)\ge\dfrac49$.

First an equivalence whose proof is immediate $$f(1)\ge\dfrac49\iff ab-a-b+\dfrac59\ge0 \ \ \ \ \ (*)$$

Also if $f(1)\ge\dfrac49$ this implies by convexity that the point $(1,\frac49)$ is outside the parabola and consequently there is the tangent of positive slope that passes through $(1,\frac49)$ and touches the parabola at a point of abscissa $x_0\in[\frac49,1)$.

Now this tangent, after elementary calculation have the following equation:

$$\frac{y-\dfrac49}{x-1}=2-a-b-2\sqrt{\frac59+ab-a-b}$$ Since this tangent exists if and only if $\dfrac59+ab-a-b\ge0$ we are done (because of the equivalence $(*)$ above).