"Prove that |log⁡(1+x2)−log⁡(1+y2)|≤|x−y|" was answered by our community

Frank Guyton

Answered question

2022-01-03

Prove that

| \log (1 + x^{2}) - \log (1 + y^{2}) | \leq | x - y |

Answer & Explanation

Annie Gonzalez

Beginner2022-01-04Added 41 answers

The derivative of

f (x) = \log (1 + x^{2})

f^{'} (x) = \frac{2 x}{1 + x^{2}}

and

| \frac{2 x}{1 + x^{2}} | \leq 1

because

2 | x | \leq 1 + {| x |}^{2}

since

{(1 - | x |)}^{2} \geq 0

Thus, by the mean value theorem,

\frac{\log (1 + x^{2}) - \log (1 + y^{2})}{x - y} = f^{'} (z)

for

z

between

x

and

y

(assuming

x \neq y

or the inequality is obvious). Then

| \frac{\log (1 + x^{2}) - \log (1 + y^{2})}{x - y} | = | f^{'} (z) \leq 1 |

Travis Hicks

Beginner2022-01-05Added 29 answers

HINT: using the mean value theorem for the function

f (t) = \ln (1 + t^{2})

we get

f (x) - f (y) = \frac{2 t}{1 + t^{2}} (x - y)

and we have

| \frac{2 t}{1 + t^{2}} | \leq 1

Vasquez

Expert2022-01-11Added 669 answers

We can write the term of interest as
$| \log (1 + x^{2}) = \log (1 + y^{2}) | = | \int_{y}^{x} \frac{2 t}{1 + t^{2}} d t |$
Now, the integrand $f (t) = \frac{2 t}{1 + t^{2}}$ attains its maximum value of 1 when t=1. Therefore, the inequality follows immediately from the mean value theorem.

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