# Proving that $f(x) = \log(1 +x^2)$ is less than $|f'(x)|$ in a small neighbourhood around $0$

plot

I am trying to prove an inequality connecting the function $$f(x) = \log(1+x^2)$$ and the absolute value of its derivative i.e $$|f'(x)| =\big|\frac{(2x)}{(1+x^2)}\big|$$ . It can be observed from the above plot of $$f(x)$$ and $$|f'(x)|$$ that $$f(x) \leq |f'(x)|$$ for $$-1 \leq x \leq1$$. I am trying to prove this fact analytically using inequalities. I have tried a couple of inequalities I could find on $$log(1+x)$$ function, but none of them is working. I also looked at some standard inequalities and series involving $$\log(x)$$, but have failed to produce any result. I am looking for directions to proceed with this. Any useful inequalities I can use to prove this would be appreciated. The plot is the only source i have in showing that this inequality is true.

EDIT : Thank you for all the kind responses. I was wondering if the proof given below are extendable to prove the following 2 extensions. Note : I am in the process of working them out. I will update here if i get these proofs. It looks like we can prove these too.

1. Can we prove that $$f(x) \leq |f'(x)|^2$$. Here we are having derivative magnitude square instead of just absolute value. Graphically it is true for $$|x|\leq 1$$
1. Can we prove that if $$f(x,y) = log (1+x^2 +y^2)$$ then $$f(x,y) \leq \lVert \nabla f(x,y)\rVert_2^2$$ in some $$\epsilon$$ neighbourhood around $$[0,0]^T$$
• @JohnHughes Sorry, i was confused as to which is the better thing to do. Post a new question or extend the question. I did not choose to do the first as I thought it could be seen as duplication. I am new here. I think you have a point there. I apologize. I will do as instructed next time. – Engineer_2018 Oct 23 '18 at 12:07
• I could see (from your reputation) that you're new here, and I merely wanted to advise you on how folks generally do things here. No one expects new folks to know how everything works, so what you've done here is just fine for a first question, and there's no need to apologize. Besides, now you know how to edit your question, which lots of folks take a long time to learn. :) So: no harm done, and you've learned something. Welcome to MSE! (And thanks for asking a sort of interesting question, too!) – John Hughes Oct 23 '18 at 12:12
• Thanks. I did learn something. Have a good day! – Engineer_2018 Oct 23 '18 at 12:19

Clearly, both functions are even, so we can safely prove your point for positive values of $$x$$ only. We need to prove that there exists some $$\epsilon >0$$ such that $$\log(1+x^2) < \left|\frac{2x}{1+x^2}\right|$$ for $$x\in(0,\epsilon)$$.

First of all, since $$x>0$$, we have $$\left|\frac{2x}{1+x^2}\right| = \frac{2x}{1+x^2}$$ so that's already a simpler inequality. Second, we can safely assume that $$x<1$$ and therefore, $$\frac{2x}{1+x^2} > \frac{2x}{1+1} = x$$

Third, we can see that the derivative of the function $$f(x)=\log(1+x^2)$$ is $$0$$ at $$x=0$$, while the derivative of $$g(x)=x$$ is $$1$$ at $$x=0$$, which means that there must exist some $$\epsilon'>0$$ for which $$f(x) for all $$x\in(0,\epsilon')$$.

Therefore, for $$\epsilon = \min\{\epsilon', 1\}$$, we have

$$f(x) < g(x)=x < \frac{2x}{1+x^2} = \left|\frac{2x}{1+x^2}\right| = |f'(x)|$$ for all $$x\in(0,\epsilon)$$, ant our claim is proven.

• Thank you for your kind response. It was very helpful. Regarding your third point, is it a standard result that if $f'(x) < g'(x)$ , then there exists an $\epsilon'$ such that $f(x) < g(x)$ for $x \in (0, \epsilon')$ – Engineer_2018 Oct 23 '18 at 10:43
• @Engineer_2018 If $f(x_0)=g(x_0)$ and $f'(x)<g'(x)$, then for values of $x$ that are close to, but above $x$, you have $f(x)<g(x)$. This is a direct consequence of the fact that $f(x)=f(x_0) + f'(x_0)\cdot (x-x_0) + o(x)$. – 5xum Oct 23 '18 at 12:15
• Now i understand. Thank you – Engineer_2018 Oct 30 '18 at 10:00

HINT

Recall that we have

$$\log(1 +x^2)< x^2$$

therefore it suffices to prove that for $$x$$ positive sufficiently small

$$x^2<\frac{2x}{1+x^2} \iff x<\frac{2}{1+x^2}$$

Refer also to the related:

• Thank you sir. I am trying to proceed this way. I will update as soon as i crack the proof of the inequality $x \lt \frac{2}{1+x^2}$ – Engineer_2018 Oct 23 '18 at 11:22

$$log (1+x^{2}) \leq x^{2}$$ so all you need is $$x^{2} \leq \frac {2|x|} {1+x^{2}}$$ which is true if $$|x| \leq 1$$. We also get $$log (1+x^{2}) \leq |f'(x)|^{2}$$ by almost the same argument. Finally, the statement about two variable case is also true and the inequality holds for $$|x| \leq 1$$ and $$|y| \leq 1$$. Here also you can start with $$\log (1+x^{2}+y^{2}) \leq x^{2}+y^{2}$$ and the rest is straightforward.

• Thank You so much . Infact , I had come close to the same proof. but is stuck at exactly this point. How do we actually show that the inequality $x^2 \leq\frac{2|x|}{1+x^2}$ is true for $|x| \leq 1$ – Engineer_2018 Oct 23 '18 at 11:19
• What you need is $|x|(1+x^{2}) \leq 2$. This true because $|x| \leq 1$ and $1+x^{2} \leq 1+1=2$ – Kavi Rama Murthy Oct 23 '18 at 11:45
• Ok that is good.Thank you sir. I manged to reach the same conclusion a couple of seconds ago. Now I am looking at the case of $f(x) \leq |f'(x)|^2$ and the multidimensional case as given in the edited question. For the multidimensional case I am looking for multidimensional version of the $log(1+x^2) \leq x^2$ inequality – Engineer_2018 Oct 23 '18 at 11:52
• @Engineer_2018 I have edited the answer to include all the cases you have mentioned. – Kavi Rama Murthy Oct 23 '18 at 12:00
• Thank you sir. It is good that the inequality extends easily to higher dimensions. – Engineer_2018 Oct 30 '18 at 10:05