Let $f:[0,1]\to\mathbb R$ be defined as $$f(x)=\max\left\{\frac{|x-y|}{x+y+1}: 0\le y\le 1\right\}\text{ for }0\le x\le 1$$ Then which of the following statements is correct?
- $f$ is strictly increasing on $[0,1/2]$ and strictly decreasing on $[1/2,1]$
- $f$ is strictly decreasing on $[0,1/2]$ and strictly increasing on $[1/2,1]$
- $f$ is strictly increasing on $[0,(\sqrt{3}-1)/2]$ and strictly decreasing on $[(\sqrt{3}-1)/2,1]$
- $f$ is strictly decreasing on $[0,(\sqrt{3}-1)/2]$ and strictly increasing on $[(\sqrt{3}-1)/2,1]$
This problem requires more work than you expected!
Extremal problems with parameters are tricky, since not only the value of the maximum, but also the "morphology of the extremal configuration" depends on the parameter. In the case at hand I propose to introduce a function $g$ of two variables as follows: $$g_x(y):=\left\{\eqalign{{x-y\over x+y+1}\qquad&(0\leq y\leq x)\cr {y-x\over x+y+1}\qquad&(x\leq y\leq 1)\cr}\right.$$ Now consider $x\in[0,1]$ as fixed. Then analyze the graph of the function $y\mapsto g_x(y)$ in the $y$-interval $[0,x]$, where $g_x(y)={x-y\over x+y+1}$, and similarly analyze the graph of the function $y\mapsto g_x(y)$ in the $y$-interval $[x,1]$, where $g_x(y)={x-y\over x+y+1}$. Now you have a complete overview over the function $y\mapsto g_x(y)$ for this particular $x$. There will be a maximum value, and this is your $f(x)$ for that particular $x$.
Note that $y\mapsto g_x(y)$ is monotone on the subintervals $[0,x]$ and $[x,1]$. It follows that $$f(x)=\max\bigl\{g_x(0), g_x(x),g_x(1)\bigr\}=\max\left\{{x\over x+1}, 0,{1-x\over x+2}\right\}\ .$$