Find the maximum value of the function $$f(x)=\frac{1}{1+\left|x\right|}+\frac{1}{1+\left|x-1\right|}$$
For $1<x$, when $x$ increase both fractions decrease hence $f(x)$ decrease. Similarly for $x<0$ When $x$ decrease $f(x)$ decrease. So $f(x)$ has maximum for some $0\le x_0\le1$. We have $f(0)=f(1)=\frac32$ and $f(\frac12)=\frac43$ So I think Maximum value of $f(x)$ is $\frac32$ but not sure how to prove it. I tried the substitution $u=x-\frac12$,
$$f(u)=\frac{1}{1+|\frac12+u|}+\frac{1}{1+|\frac12-u|}$$And conclude $f(x)$ is symmetric along $x=\frac12$ but I don't know how to continue.