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

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.
