Little help proving an inequality?

Question

abs(1/(x^2+1) - 1/(y^2+1) )< abs(x - y)   (x and y in [0, infinity)

Working with the right hand side, using the some LCD's, factoring, and properties of the absolute value, etc, etc...

I work my way down to this.

|x-y||x+y|/(|x^2+1||y^2+1|)<|x-y|  So, obviously, I have the |x-y| I'm looking for on the left, so all I need to do is show that  |x+y|/(|x^2+1||y^2+1|) < 1, which by inspection sure looks like it will always be less than one on the interval given, but I can't seem to prove it.

Daniel Lopez Aguayo · Accepted Answer

Recall an everywhere differentiable map f: R-> R is Lipschitz continuous if and only if its first derivative is bounded.

The statement is basically asking you to show that f(x) = 1/(x^2+1) is Lipschitz continuous on [0,infinity).

Note that |f ' (x)| = 2|x|/(x^2+1)^2 <1 on [0,infinity) so the mean-value theorem implies that:

|f(x) - f(y)| < |x-y| for all x,y in [0,infinity), that is:

|1/(x^2+1) - 1/(y^2+1) | < |x-y| for all x,y in [0,infinity), as desired.

In fact using derivatives one can show more:

|1/(x^2+1) - 1/(y^2+1)| <= (3√3/8) |x-y| for all x,y in [0,infinity)

i.e f is Lipschitz continuous on [0,infinity) with constant 3√3 / 8, which is less than 1 so this sharpens the original inequality.

vekkus4 · Answer

x and y in [0, infinity) so every term here is greater than zero, it will be simpler if the absolute values aren't used

To prove |x+y| < |x^2 + 1||y^2 + 1|

x + y < x^2 y^2 + x^2 + y^2 + 1

It is easily true for x,y >= 1 because x^2 > x and y^2 > 1

If x+y < 1 then it's easily true because all the square terms are positive

If y>=1 it suffices to prove x < x^2 + 1 because y^2 >= 1. That reduces to proving x^2 - x + 1 > 0 so complete the square - (x - 1/2)^2 + 3/4 > 0 is true. Similarly if x >= 1 it is true.

Now what remains is a small triangle where x+y > 1 but x<1 and y < 1

I think it works nicely like this

x+y > 1 => (x+y)^2 > (x+y)

But (1 + x^2)(1+y^2) > (x+y)^2

because

1 + x^2 + y^2 + x^2 y^2 > x^2 + y^2 + 2 x y

because 1 + x^2 y^2 > 2 x y

because (1 + xy)^2 > 0

Siths and Giggles · Answer

| 1/(x²+1) + 1/(y²+1) | < |x - y|

| ((y² + 1) - (x² + 1)) / ((x² + 1)(y² + 1)) | < |x - y|

| (y² - x²) / ((x²+1)(y²+1)) | < |x - y|

| ((y - x)(y + x)) / ((x²+1)(y²+1)) | < |x - y|

| (-(x - y)(y + x)) / ((x²+1)(y²+1)) | < |x - y|

(|x - y| |y + x|) / (|x²+1| |y²+1|) < |x - y|

I'm not sure if it would be valid to divide both sides by |x - y|, considering x and y can both be 0 judging by the interval you've been given, but for the case (0, ∞):

|y + x| / (|x²+1| |y²+1|) < 1
|y + x| < |x²+1| |y²+1|

Have you tried applying the Triangle Inequality?
|x+y| ≤ |x| + |y|

|x| + |y| < |x²+1| |y²+1|,
which does work ∀ x,y ∈ (0,∞)

As you can probably tell, my knowledge of mathematical proofs is limited.

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Little help proving an inequality?

3 Answers