Prove $|f(y) − f(x)| \leq f(|y − x|)$ if $|y − x| ≤ 1/2$ given $f(x)=-x\log_2 x$ 
How do I prove that whenever $|y − x| ≤ 1/2$ it follows that $|f(y) − f(x)| \leq f(|y − x|)$ given $f(x)=-x\log_2 x$?

where $x,y\in[0,1]$
The graph of $f(x)=-x\log_2 x$ function is

$$
f'(x)=-\log x-\frac{x}{x\ln 2}=-\log x-\frac{1}{\ln 2}=0\implies \log x=-\frac{1}{\ln 2}\\
\frac{\ln x}{\ln 2}=-\frac{1}{\ln 2}\implies \ln x=-1\implies x=1/e\approx 0.3679
$$
I was only able to write the following proof:
When $0\le x\le y\le 1$,
\begin{align}
|f(x)-f(y)|&=|-x\log x+y\log y|\\
&=|-x\log x+\frac{x}{\ln 2}+y\log {y}-\frac{y}{\ln 2}+\frac{y}{\ln 2}-\frac{x}{\ln 2}|\\
&\leq |-(x\log x-\frac{x}{\ln 2})+(y\log y-\frac{y}{\ln 2})|+\frac{|y-x|}{\ln 2}\\
&=|\int_x^y \log tdt|+\frac{|y-x|}{\ln 2}=-\int_x^y \log tdt+\frac{|y-x|}{\ln 2}\\
&=-\int_x^{x+(y-x)} \log tdt+\frac{|y-x|}{\ln 2}\\
&\leq -\int_0^{y-x} \log tdt+\frac{|y-x|}{\ln 2}\\
&=-\Big(t\log t-t\Big)_0^{y-x}+\frac{|y-x|}{\ln 2}\\
&=-(y-x)\log(y-x)+(y-x)+\frac{|y-x|}{\ln 2}\\
&=f(|y-x|)+|y-x|+\frac{|y-x|}{\ln 2}
\end{align}
My Attempt
Thanks @Balajisb for the hint.
If $f$ is a concave function then $f(a+b)\leq f(a)+f(b)$ for all $a,b>0$
$$
-y\log y=-(y-x+x)\log(y-x+x)\le-(y-x)\log(y-x)-x\log x\\
-y\log y-(-x\log x)\le -(y-x)\log(y-x)\\
f(y)-f(x)\le f(y-x)
$$
$$
D_{\log x}\in(0,\infty]\implies x,y,y-x\geq 0\implies 1\ge y>x>0\\
$$
$$
1>y-x>0\implies -(y-x)\log(y-x)>0
$$
$$
f(y)-f(x)\le f(|y-x|) \;\forall\;(x,y)\;|\;y>x>0\\f(x)-f(y)\le f(|y-x|) \;\forall\;(x,y)\;|\;x>y>0
$$
Therefore,
$$
|f(y)-f(x)|\le f(|y-x|) \;\forall\;(x,y)\;|\;y>x>0\;\&\;f(y)>f(x)\\|f(y)-f(x)|\le f(|y-x|) \;\forall\;(x,y)\;|\;y<x<0\;\&\;f(y)<f(x)
$$
In order to prove $|f(y)-f(x)|\le f(|y-x|) \;\forall\;x,y>0$ we need to also consider the cases $y>x>0\;\&\;f(y)<f(x)$ and $x>y>0\;\&\;f(x)<f(y)$. So I think that's where the condition $y-x\leq 1/2$ lies in.
Case 1 : $y>x>0\;\&\;f(y)<f(x)$
$$
-y\log y<-x\log x\implies y\log y>x\log x\\
x\log x-y\log y<0\\x\log x-\frac{x}{\ln 2}-y\log y+\frac{y}{\ln 2}-\frac{y-x}{\ln 2}<0\\
\int_y^x \log t dt-\frac{y-x}{\ln 2}>0\\
y-x<\ln 2\int_y^x \log t dt=-\ln 2\int_x^y \log t dt=-\ln 2\int_x^y \frac{\ln t}{\ln 2}dt=-\int_x^y \ln t dt\\
<-\int_0^1 \log t dt=-1\times -1=1
$$

How do I obtain the condition $y-x\leq 1/2$ in this case ?


Assume without loss of generality that $ y > x$, and write $z = y - x$, so that $0 \leq  x \leq 1-z \lt 1 \ \text{and} \  z \leq 1/2$ .
Then the concavity of $f$ tells us that the slope of the chord joining $(x,f(x))$ and $(x+z,f(x+z))$ exceeds the slope of the chord joining $(1-z,f(1-z)) $ and $(1,f(1))$, which gives us $f(y) - f(x) \geq -f(1-z)$.
Likewise the slope of the chord joining $(x,f(x))$ and $(x+z,f(x+z))$ is smaller than the slope of the chord joining $(0,f(0)) $ and $(z,f(z))$, which gives $f(y) - f(x) \leq f(z)$.
Hence, $|f(y) - f(x)| \leq \max\{f(z),f(1-z) \} = f(z)$, as $z \leq 1/2$.
Note : If $f$ is convex, and $ a \leq b ;c \leq d $, where $ a \leq c $ and $ b \leq d $, then the slope of the chord joining $(b,f(b)), (d,f(d))$ exceeds that of the chord joining $(a,f(a)), (c,f(c))$.
