 Solution

ID: 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/solution

x

and

a / x

gives

g (x) = \frac{1}{2} (x + \frac{a}{x}) \geq a .

Equality holds precisely when

x = a / x

, hence at

x = a

. Thus

x > 0 min g (x) = a

, attained only at

x = a

. Equivalently, the exact identity

g (x) - a = (x - a)^{2} / (2 x)

proves the same global bound.

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