Prove $\arctan(x+y)0$ Prove $\arctan(x+y)<y+\arctan(x),x\in \mathbb{R}, y>0$
Using Lagrange's mean value theorem, 
$$f(u)=\arctan(u)$$ 
In the first case, $x>0$
$$x=a,x+y=b$$
$f(u)$ is continuous and differentiate on $(a,b)$,
$$\exists c \in (a,b): \;\left(\frac{f(b)-f(a)}{b-a}=f^{'}(c)\right)$$
$$\frac{\arctan(b)-\arctan(a)}{y}=\frac{1}{1+c^2}$$
$$y=(1+c^2)(\arctan(b)-\arctan(a))$$
$$\arctan(b)<\arctan(a)+(1+c^2)(\arctan(b)-\arctan(a))$$
If $c\in (a,b)$ then inequality is true.
In the second case, $x=0$
$$x=0,b=y$$
$$\frac{\arctan(b)}{y}=\frac{1}{1+c^2}$$
$$y=(1+c^2)\arctan(b)$$
$$\arctan(b)<(1+c^2)\arctan(b)$$
Inequality is true.
In the third case, $x<0$, inequality is true (first case).
Is this proof correct?
Could someone check?
 A: Another way to get this is to note
$$\arctan(x+y)-\arctan x = \int_x^{x+y}\frac{dt}{1+t^2}.$$
The integrand is less than $1$ on $[x,x+y]$ except for at most one point. Therefore the integral is less than the length of $[x,x+y]$ times $1,$ i.e., less than $y\cdot 1 = y.$
A: You get from mean value theorem that $$\frac{\arctan(x+y)-\arctan x}y=\frac{1}{1+c^2}$$
for some $c\in(x,x+y)$, hence (using $y>0$)
$$\arctan(x+y)-\arctan x=\frac y{1+c^2}\ge y $$
with equality only if $c=0$. However, if $c=0$ then $x<c<x+y$ and then by cases where we have strict inequality because $0$ is not in the interval
$$\begin{align}\arctan(x+y)-\arctan x&=(\arctan(x+y)-\arctan 0)+(\arctan 0-\arctan x)\\&>(x+y-0)+(0-x)=y \end{align}$$
If you think about it, this works for any differentiyble function $f$ for which $f'(x)>1$ holds for $x$ in a dense subset.
A: There is a mistake in your proof, you can't infer the inequality $\arctan(b) < \arctan(a) + (1 + c^2)(\arctan(b) - \arctan(a))$, since you've established equality of these terms just two lines earlier. Also, the different cases aren't necessary at all.
