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We'll write the inequality:
(x-1)/x < ln x
We'll subtract ln x both sides:
(x-1)/x - ln x < 0
We'll assign a function to the expression above:
f(x) = (x-1)/x - ln x
This function is differentiable over the range [1 ; +infinite)
f'(x) = 1/x^2 - 1/x = (1 - x)/x^2
Since the values of x are in the range [1 ; +infinite), then the numerator of the fraction (1 - x)/x^2 is always negative, for any value of x. Since the denominator is always positive, since it is a square, then the fraction is negative for any value of x.
Therefore, the function is decreasing.
So, for x >= 1 => f(x) < f(1)=0
But f(x) = (x-1)/x - ln x => (x-1)/x - ln x < 0 => (x-1)/x < ln x
The inequality (x-1)/x < ln x is verified, for all real values of x, located in the range [1 ; +infinite).
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