Past exam of the year 1 of the computer science course of the University of Oxford / 2014 / Discrete mathematics, continuous mathematics and probability / 5 / a

The arithmetic-geometric mean inequality for the positive numbers $x$ and $a/x$ givesEquality holds precisely when $x=a/x$, hence at $x=\sqrt{a}$. Thus $\boxed{\underset{x>0}{\min }g(x)=\sqrt{a}}$, attained only at $\boxed{x=\sqrt{a}}$. Equivalently, the exact identity $g(x)-\sqrt{a}=(x-\sqrt{a}{)}^2/(2x)$ proves the same global bound.