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 

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: $f'(0)=\pm\frac1{(1+0)^2}-\frac1{(1+a)^2}$.
