Local maxima and minima of $f(x)=\frac{1}{1+|x|}+\frac{1}{1+|x-a|}$ 
Local $\max$ and local $\min$ value of $$f(x)=\frac{1}{1+|x|}+\frac{1}{1+|x-a|}\;,$$ Where $a>0$

$\bf{My\; Try::}$ $\bf{For\; maximum}$
Using $$1+\frac{1}{1+|a|}-f(x)=1+\frac{1}{1+|a|}-\frac{1}{1+|x|}-\frac{1}{1+|x-a|}$$
$$ = \frac{|x|}{1+|x|}+\frac{|x-a|-|a|}{\left(1+|a|\right)(1+|x-a|)}\geq \frac{|x|}{1+|x|}+\frac{|a|-|x|-|a|}{\left(1+|a|\right)(1+|x-a|)}$$
$$ =|x|\bigg[\frac{1+|a|+|x-a|+|a||x-a|-1-|x|}{(1+|x|)\left(1+|a|\right)(1+|x-a|)}\bigg]\geq |x|\bigg(\frac{|a||x-a|}{(1+|x|)\left(1+|a|\right)(1+|x-a|)}\bigg)\geq 0$$
So $$f(x)\leq 1+\frac{1}{1+|a|}$$
and equality hold when $x=0$ and $x=a$
But i did not understand how can i calculate $\min,$ Help required, Thanks
 A: Note that $f$ is symmetric around $x=a/2$. Although $f$ has no global minimum but you can easily verify that there is a local minimum at $x=a/2$.
Here's my proof, however, I know the question owner actually doesn't need any more details to proceed. Assuming $0<\delta<a/2$, we're going to show $f(a/2)< f(a/2+\delta)$ and then due to symmetry, $f(a/2)< f(a/2\pm\delta)$ will be concluded. First, for $f(a/2)$ we know
\begin{aligned}
f\left(\frac a2\right)&=2\frac1{1+\frac a2}=\frac 4{2+a}
\end{aligned}
Now for $f(a/2+\delta)$
\begin{aligned}
f\left(\frac a2+\delta\right)&=\frac1{1+\frac a2+\delta}+\frac1{1+\frac a2-\delta}\\
&=\frac{2+a}{\left(1+\frac a2\right)^2-\delta^2}\\
&= \frac{4(2+a)}{(2+a)^2-4\delta^2}\\ 
&=\left(\frac {2+a}4 -\frac {\delta^2}{2+a}\right)^{-1}\\
&=\left(f\left(\frac a2\right)^{-1}-\epsilon\right)^{-1},\quad \epsilon=\frac {\delta^2}{2+a}>0
\end{aligned}
This implies
$$f\left(\frac a2\right)<f\left(\frac a2+\delta\right)$$ And due to symmetry $$f\left(\frac a2\right)<f\left(\frac a2\pm\delta\right)$$ Thus, $f$ has a local minimum at $x=a/2$.
