# Prove that 1 - 2x/(x^2+1)>=0

Asked on by ggenius

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to prove that 1 - 2x / (x^2 + 1) >= 0

(1 - x)^2 >= 0

=> 1 - 2x + x^2 >= 0

divide by (1 + x^2). As (1 + x^2)  is always positive the inequality holds.

=> (1 + x^2)/(1 + x^2) - 2x/(1 + x^2) >=0

=> 1 - 2x/(1 + x^2) >= 0

This proves that 1 - 2x / (x^2 + 1) >= 0

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll multiply 1 by (x^2 + 1) and we'll get:

(x^2 + 1 - 2x)/(x^2 + 1) >=0

A fraction is positive when the numerator and denominator are both positive or both negative.

The numerator of the fraction is the perfect square (x-1)^2 that is positive for any real x, except x = 1, for the numerator is cancelling out.

The denominator is always positive, for any real value of x.

Therefore, the inequality 1 - 2x/(x^2+1) = (x-1)^2/(x^2 + 1) >=0 is verified for any real x.

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