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The exercise is:

Show, using Lagrange's theorem, that for $x \in [0, +\infty] $, we have $ \frac{x}{1+x^2} \leq \arctan(x)$.

I know how to apply Lagrange's theorem but my trouble is to find a function to apply it.

I thought about $f(x)= \arctan (x) $ but i doesn't seem to be working.

Can someone give a hint please?

Thanks!

Mark Viola
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Granger Obliviate
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2 Answers2

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Let $f(x)=\arctan(x)-\frac{x}{1+x^2}$. Note that $f(0)=0$ and

$$f'(x)=\frac{2x^2}{(1+x^2)^2}\ge 0$$

From the mean-value theorem, there exists a number $\xi\in (0,x)$ such that

$$\begin{align} f(x)&=f(0)+f'(\xi)x\\\\ &=\frac{2\xi^2}{(1+\xi^2)^2}x\\\\ &\ge 0 \end{align}$$

And we are done!

I thought that it might be instructive to present an approach that does not rely on differential calculus, but rather on an elementary inequalities from geometry. In THIS ANSWER, I showed using only the inequalities

$$x\cos(x)\le \sin(x)\le x$$

for $x>0$, that the arctangent function satisfies the inequalities

$$\frac{x}{\sqrt{1+x^2}}\le \arctan(x)\le x \tag 1$$

Since $\frac{1}{\sqrt{1+x^2}}\ge \frac{1}{1+x^2}$, then we also see from $(1)$ that

$$\arctan(x)\ge \frac{x}{1+x^2} \tag 2$$

for $x>0$. Therefore, $(1)$ actually provides a tighter lower bound than $(2)$ for the arctangent.

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Mark Viola
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  • First of all thanks for the reply! Second, where are you applying Lagrange theorem? – Granger Obliviate Apr 07 '16 at 17:29
  • You're welcome. My pleasure. Are you not referring to the MVT? – Mark Viola Apr 07 '16 at 17:32
  • There are six different theorems that are regularly called "Lagrange's Theorem"! Apparently you are referring to the 'mean value theorem' that is due to Lagrange but not often referred to as "Lagrange's Theorem". – user247327 Apr 07 '16 at 17:47
  • Granger, please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark – Mark Viola May 09 '16 at 16:56
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Take your function $f(x)=\arctan(x)$. The derivative is $\frac{1}{1+x^2}$.

Now use Lagrange theorem for two points $0,x$ when $x>0$.

$$\frac{\arctan(x)-\arctan(0)}{x-0}=\frac{1}{1+x_0^2}$$

$x_0$ is a real number between $0$ and $x$.

By rewriting the last result, we get:

$\arctan(x)-\arctan(0)=\frac{x}{1+x_0^2}$

It is time to turn the equality into an inequality by replacing $x_0$ with $x$. Knowing that $x>x_0$, we have reduced the value of the fraction by doing the replacement and consequently we get:

$\arctan(x)\geqslant\frac{x}{1+x^2}$

Mark Viola
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