We'll apply calculus, namely Lagrange's theorem, to prove the given inequality.

We'll choose a function, whose domain of definition is the closed interval [a,b].

The function is f(x) = ln x

Based on Lagrange's theorem, there is a point "c", that belongs to (a,b), so that:

f(b) - f(a) = f'(c)(b - a)

We'll substitute the function f(x) in the relation above:

ln b - ln a = f'(c)(b-a)

We'll determine f'(x):

f'(x) = (ln x)'

f'(x) = 1/x

f'(c) = 1/c

ln b - ln a = (b-a)/c

ln (b/a) = (b-a)/c

Since c is in the interval [a,b], we'll get the inequality:

a<c<b => 1/a > 1/c > 1/b

We'll multiply by the positive amount (b-a):

(b-a)/a > (b-a)/c > (b-a)/b (1)

But (b-a)/c = ln (b/a) (2)

We'll substitute (2) in (1) and we'll get the inequality that has to be demonstrated:

**(b-a)/a > ln (b/a) > (b-a)/b**