Not sure if this is in calculus section as I am not sure of approach.

Given

f(x) = (a+b+x)/(3abx) , x>0 and a ,b are positive real numbers

i)Show that the minimum value of f(x) occurs when x = (a + b)/2

Thanks. No idea how to approach. Tried differentiation but ended with

f'(x) = -(a+b)/(3abx^2) which does not have turning point.