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

Question

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.

Answer 1

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$.

Answer 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 $f(x)$.