Let f x = x - a x - b - a + b 2 . If f x = 0 has both non-negative roots, then the minimum value of f x

Given,

$f\left(x\right)=\left(x-a\right)\left(x-b\right)-\left(\frac{a+b}{2}\right)$

$\Rightarrow f\left(x\right)=x^2-\left(a+b\right)x-\left(\frac{a+b}{2}\right)$

Let $\alpha \& \beta$ be the roots of $f\left(x\right)$.

$\alpha , \beta >0$ where the minimum is obtained i.e. $\frac{-b}{2a}>0$

$\frac{a+b}{2}>0\Rightarrow a+b>0 ...\left(i\right)$

Now, 

$x^2-\left(a+b\right)x-\left(\frac{a+b}{2}\right)=0$

$\Rightarrow x=\frac{\left(a+b\right)\pm \sqrt{{\left(a+b\right)}^2-4\left(ab+{\displaystyle \frac{a+b}{2}}\right)}}{2}$

Roots are non-negative, so

$\frac{\left(a+b\right)\pm \sqrt{{\left(a-b\right)}^2+2\left(a+b\right)}}{2}\ge 0$

$\Rightarrow \left(a+b\right)\pm \sqrt{{\left(a-b\right)}^2+2\left(a+b\right)}\ge 0$

$\Rightarrow {\left(a+b\right)}^2\ge {\left(a-b\right)}^2+2\left(a+b\right)$

$\Rightarrow \left(a+b\right)\le 2ab ...\left(ii\right)$

By $\left(i\right) \& \left(ii\right)$, we have

$0<\left(a+b\right)\le 2ab$

Minimum value$=\frac{-D}{4a}$

 $=-\left\{\frac{{\left(a-b\right)}^2+2\left(a+b\right)}{4}\right\}$

   $=-\left\{\frac{{\left(a-b\right)}^2+4ab}{4}\right\}$

   $=-\left\{\frac{{\left(a+b\right)}^2}{4}\right\}$

Therefore, $f\left(x\right)\ge -\frac{{\left(a+b\right)}^2}{4}$.