Proof with Lagrange theorem 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! 
 A: 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.
A: 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:
$\arctan(x)\geqslant\frac{x}{1+x^2}$
