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

Question

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 $|x|\leq 1$
   2. 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 $[0,0]^T$

Answer 1

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:

• Show these simple inequalities and $(\log(1+x))^2\le x$ and that $(\log(1+x))^2\le x^2$

Answer 2

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 $f(x)<g(x)$ 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.

Answer 3

$ 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.