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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
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