<s>[INST] 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)
 [/INST] 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 
$$f(1)=2f(\frac12)+\frac12-ab\ge \frac2{36}+\frac12 - \frac19 = \frac49
\implies f(1) \ge \frac49$$
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