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 ?


Here is a proof.
WLOG, assume that $x \le y$.
We need to prove that
$$|-y\ln y + x\ln x| \le -(y - x)\ln(y - x). \tag{1}$$
Using the identity for $u \ge 0$ (easy to prove)
$$u\ln u =\int_0^1 \frac{u(u - 1)}{1+(u-1)t}\,\mathrm{d} t,$$
(1) is written as
$$\left|\int_0^1 \left(\frac{y(1-y)}{1+(y-1)t} - \frac{x(1-x)}{1+(x-1)t}\right)\,\mathrm{d} t\right|
\le \int_0^1 \frac{(y-x)(1 - (y-x))}{1 + (y-x - 1)t}\,\mathrm{d} t.$$
It suffices to prove that
$$\int_0^1 \left|\frac{y(1-y)}{1+(y-1)t} - \frac{x(1-x)}{1+(x-1)t}\right|\,\mathrm{d} t
\le \int_0^1 \frac{(y-x)(1 - (y-x))}{1 + (y-x - 1)t}\,\mathrm{d} t.$$
It suffices to prove that, for all $t \in [0, 1]$ and $0 \le x \le y \le 1$ with $y - x \le 1/2$,
$$\left|\frac{y(1-y)}{1+(y-1)t} - \frac{x(1-x)}{1+(x-1)t}\right|
\le \frac{(y-x)(1 - (y-x))}{1 + (y-x - 1)t}$$
or
$$\left(\frac{y(1-y)}{1+(y-1)t} - \frac{x(1-x)}{1+(x-1)t}\right)^2
\le \left(\frac{(y-x)(1 - (y-x))}{1 + (y-x - 1)t}\right)^2$$
or (clearing the denominators and simplifying)
$$xy[1 - 2(y - x)]t^2 + (2x^2 - xy - 2y^2 + 2y)t(1 - t) + 2(1-y)(1-t)^2 \ge 0 \tag{2}$$
which is true (the proof is given at the end).
We are done.

Proof of (2):
It suffices to prove that $2x^2 - xy - 2y^2 + 2y \ge 0$.
If $0 \le x \le 1/3$, then
$$2x^2 - xy - 2y^2 + 2y
\ge 2x^2 - xy - 2y\cdot (x + 1/2) + 2y = 2x^2 + y(1 - 3x) \ge 0.$$
If $1/3 < x < 1/2$, then
$$2x^2 - xy - 2y^2 + 2y
\ge 2x^2 - xy - 2y\cdot (x + 1/2) + 2y = 2x^2 - y(3x - 1) $$
$$\ge 2x^2 - (x + 1/2)(3x - 1) = \frac12(1 + x)(1 - 2x) \ge 0.$$
If $1/2 \le x \le 1$, then $2x^2 - xy - 2y^2 + 2y
= x(2x - y) + 2y(1 - y) \ge 0$.
We are done.
