# Maximizing $f(x)=\frac{1}{1+\left|x\right|}+\frac{1}{1+\left|x-1\right|}$

Find the maximum value of the function $$f(x)=\frac{1}{1+\left|x\right|}+\frac{1}{1+\left|x-1\right|}$$

For $$1, when $$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.

You are almost done. In $$[0,1]$$ check that the derivative is positive for $$x <\frac 1 2$$ and negative for $$x>\frac 1 2$$. This proves that the maximum value in $$[0,1]$$ is attained at $$\frac 1 2$$ (since $$f$$ is increasing in $$[0,\frac 1 2]$$ and decreasing in $$[\frac 1 2, 1]$$). Hence, the maximum value of $$f$$ is indeed $$\frac 3 2$$.
I just solved the problem! Since for $$0\le x_0\le1$$ we have maximum, we can get rid of absolute bars,
$$f(x)=\frac1{x+1}+\frac{1}{2-x}=\frac3{(x+1)(2-x)}$$ Denominator is a quadratic and it is maximum for $$x=\frac12$$ and for $$x\in[0,1]$$ for either $$x=0$$ or $$x=1$$ it is minimum. Hence $$x=0$$ or $$x=1$$ maximize $$f(x)$$.