Monotonicity of $f(x)=\max\left\{\frac{|x-y|}{x+y+1}:0\le y\le1\right\}$ for $x\in[0,1]$ Let the function $f:[0,1] \rightarrow \mathbb{R}$ be defined as
$$f(x)=\max\left\{\frac{|x-y|}{x+y+1}:0\le y\le1\right\}\ \ \text{ for }\ \ 0 \le x \le 1\,,$$
then which of the following statements is correct?
(A) $f$ is strictly increasing on $\left[0,\frac{1}{2}\right]$ and strictly decreasing on $\left[\frac{1}{2},1\right]$
(B) $f$ is strictly decreasing on $\left[0,\frac{1}{2}\right]$ and strictly increasing on $\left[\frac{1}{2},1\right]$
(C) $f$ is strictly increasing on $\left[0,\frac{\sqrt3-1}{2}\right]$ and strictly decreasing on $\left[\frac{\sqrt3-1}{2},1\right]$
(D) $f$ is strictly decreasing on $\left[0,\frac{\sqrt3-1}{2}\right]$ and strictly increasing on $\left[\frac{\sqrt3-1}{2},1\right]$
Apparently, this question was asked in an exam for high school students and shouldn't involve multi-variable calculus. I tried differentiation, trigonometric substitution, and modulus inequalities but nothing seems to be working. My question is how to solve it without multi-variable calculus?  
 A: This solution only uses the following elementary facts.  First, for any $c>0$, the map $x\mapsto \dfrac{1}{x+c}$ is strictly decreasing for $x\geq 0$. Second, if a real-valued function $h$ defined on a subset of $\mathbb{R}$ is a strictly decreasing function, then $\lambda\,h$ is also strictly decreasing for $\lambda>0$, and $\lambda\,h$ is strictly increasing for $\lambda<0$.  Third, if a real-valued function $h$ defined on a subset of $\mathbb{R}$ is a monotonic function, then $h+c$ is also monotonic with the same monotone-type as $h$ for any $c\in\mathbb{R}$.
For a fixed $x\in[0,1]$, let $g_x:[0,1]\to\mathbb{R}$ be the function given by
$$g_x(y):=\frac{|x-y|}{x+y+1}$$
for all $y\in[0,1]$.  For $y\in[0,x]$, we have
$$g_x(y)=\frac{x-y}{x+y+1}=\frac{2x+1-(x+y+1)}{x+y+1}=\frac{2x+1}{x+y+1}-1\,.$$
As $y\mapsto \dfrac{2x+1}{x+y+1}$ is a strictly decreasing function for $y\geq 0$, this means $g_x$  is strictly decreasing on $[0,x]$.  
For $y\in[x,1]$, we have
$$g_x(y)=\frac{y-x}{x+y+1}=\frac{(x+y+1)-(2x+1)}{x+y+1}=1-\frac{2x+1}{x+y+1}\,.$$
Thus, $g_x$ is strictly increasing on $[x,1]$.  This shows that
$$f(x)=\max_{x\in[0,1]}\,g_x(y)=\max\big\{g_x(0),g_x(1)\big\}$$
for all $x\in[0,1]$.  
Ergo, for every $x\in[0,1]$, we have
$$f(x)=\max\left\{\frac{x}{x+1},\frac{1-x}{x+2}\right\}\,.$$
Because $x\mapsto \dfrac{x}{x+1}=1-\dfrac{1}{x+1}$ is strictly increasing and $x\mapsto \dfrac{1-x}{x+2}=\dfrac{3}{x+2}-1$ is strictly decreasing for all $x\geq 0$, we conclude that, if $u$ is the real number in $[0,1]$ such that 
$$\frac{u}{u+1}=\frac{1-u}{u+2}\,,$$
then
$$f(x)=\left\{\begin{array}{ll}\dfrac{1-x}{x+2}&\text{if }0\leq x\leq u\,,\\
\dfrac{x}{x+1}&\text{if }u\leq x\leq 1\,.\end{array}\right.$$
By solving for $u$, we have $(2u+1)^2=3$, so
$$u=\frac{\sqrt{3}-1}{2}\,.$$
Thus, $f$ is strictly decreasing on $\left[0,\dfrac{\sqrt3-1}{2}\right]$ and strictly increasing on $\left[\dfrac{\sqrt3-1}{2},1\right]$.
