# Verify if the ratio (x-1)/x is smaller than lnx if x>=1.

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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).**