SE-SKOLORNAS 1973 Problem 6

$f(x)$ is a real valued function defined for $x \geq 0$ such that $f(0) = 0$ , $f(x+1)=f(x)+\sqrt{x}$ for all $x$ , and $$f(x) < \frac{1}{2}f\left(x - \frac{1}{2}\right)+\frac{1}{2}f\left(x + \frac{1}{2}\right) \text{for all} x \geq \frac{1}{2}$$ Show that $f\left(\frac{1}{2}\right)$ is uniquely determined.