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)

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).
