Find the maximum of $\frac{1}{1+|x|}+\frac{1}{1+|x-a|}$ Let $a>0$. Show that the maximum value of the function 
$$f(x)= \frac{1}{1+|x|}+\frac{1}{1+|x-a|}$$
is $$\frac{2+a}{1+a}.$$
really need some help with this thing 
 A: Study what happens in each of the three regions $\{x<0\}$, $\{0\le x\le a\}$ and $\{x>a\}$.
A: $f(a) = \dfrac{2+a}{1+a}$, and $f(x) = f(a-x)$, it suffices to consider $f$ on $[0,\frac{a}{2}]$, and on this interval $f(x) = \dfrac{1}{1+x} + \dfrac{1}{1+a-x}$, and $f'(x) = -\dfrac{1}{(1+x)^2} + \dfrac{1}{(1+a-x)^2} \leq 0$, thus $f(x) \leq f(0) = \dfrac{2+a}{1+a}$.
A: The function $\color{green}{\frac1{1+|x|}}$ is even and decreases from an angular point at the origin.  $\color{magenta}{\frac1{1+|x-a|}}$ is the same function, shifted by $a$.
The $\color{blue}{\text{sum}}$ has two symmetric maxima, at the angular points $x=0$ and $x=a$, where the function equals $\frac1{1+0}+\frac1{1+a}$.

For complete rigor, one must show that the derivative changes sign at the angular points, which is always true: $f'(0)=\pm\frac1{(1+0)^2}-\frac1{(1+a)^2}$.
A: Hint: Find the derivative of $f$ using the fact that 
$$
\frac{\partial}{\partial x}|x|=\frac{x}{|x|}.
$$
Then find out where $\frac{\partial f}{\partial x}=0$, or just try to find out what is happening for the derivative when $x=\frac{2+a}{1+a}$.
