Proving that: $||x|^{s/2}-|y|^{s/2}|\le 2|x-y|^{s/2}$ Given $ 0<s<1$  and $x,y\in \Bbb R^d$, such that $|x|,|y|\ge 1$ 

I would like to prove that 
  $$||x|^{s/2}-|y|^{s/2}|\le 2|x-y|^{s/2}$$

so far I am not getting the good bound. Here is what I have tried
$$\begin{align}|x|^{s/2}&\le (|x-y|+|y|)^{s/2}\le 2^{s/2}\max\{|x-y|^{s/2},|y|^{s/2}\}\\
&\le2^{s/2+1}(|x-y|^{s/2}+|y|^{s/2})\end{align}$$
This estimate doesn't seem to be helpful.
 Any idea?
 A: Define $f(x)=(x+1)^{\frac{s}{2}}-x^{\frac{s}{2}}-1$, $x>0$. Note that, $f'(x)=\frac{s}{2}\left((x+1)^{\frac{s-2}{2}}-x^{\frac{s-2}{2}}\right)\le0$, because $x+1\ge x$ and $\frac{s-2}{2}<0.$ So, for $x>0$, $f$ is decreasing, then $f(x)\le f(0)$ for $x>0$. Therefore, 
$$(x+1)^{\frac{s}{2}}\le x^{\frac{s}{2}}+1.$$
Replace $x$ for $\frac{x}{y}$, $x,y>0$, we obtain,
$$\left(\frac{x}{y}+1\right)^{\frac{s}{2}}\le \left(\frac{x}{y}\right)^{\frac{s}{2}}+1.$$
Therefore,
$$\left(x+y\right)^{\frac{s}{2}}\le x^{\frac{s}{2}}+y^{\frac{s}{2}}\tag{1}.$$
Now using $(1)$ we have
$$
|x|^{\frac{s}{2}}\leq (|x-y|+|y|)^{\frac{s}{2}}\leq  |x-y|^{\frac{s}{2}}+|y|^{\frac{s}{2}}.
$$
Hence,
$$
|x|^{\frac{s}{2}}-|y|^{\frac{s}{2}}\leq  |x-y|^{\frac{s}{2}}.
$$
At the same way
$$
|y|^{\frac{s}{2}}\leq (|y-x|+|x|)^{\frac{s}{2}}\leq  |x-y|^{\frac{s}{2}}+|x|^{\frac{s}{2}}.
$$
Then,
$$
|y|^{\frac{s}{2}}-|x|^{\frac{s}{2}}\leq  |x-y|^{\frac{s}{2}}.
$$
Therefore
$$
||x|^{\frac{s}{2}}-|y|^{\frac{s}{2}}|\leq  |x-y|^{\frac{s}{2}}\leq2|x-y|^{\frac{s}{2}}.
$$
