Please explain inequality $|x^{p}-y^{p}| \leq |x-y|^p$

Suppose $x \geq 0$, $y \geq 0$ and $0<p<1$. Why is the following inequality true?

$|x^{p}-y^{p}| \leq |x-y|^p$

Assume w.lo.g $x>y$. Then the statement becomes $x^p - y^p \leq (x-y)^p$. Since $y > 0$, divide through out by $y^p$. So we need to show $(\frac{x}{y})^p - 1 \leq (\frac{x}{y}-1)^p$ whenever $x > y >0$ and $0<p<1$. Let $t = \frac{x}{y}$. So we need to show $t^p - 1 \leq (t-1)^p$ whenever $t > 1$ and $0<p<1$.

This is a calculus problem.

Let $f(t) = (t-1)^p - (t^p -1) $ where $0<p<1$.

Show that the function $f(t)$ is increasing for $t \geq 1$ and when $0 < p < 1$.

So $f(t) \geq f(1)$ and $f(1) = 0$. So, we get the desired result.

$\textbf{EDIT:}$ To show $f(t)$ is increasing for $t \geq 1$ and when $0 < p < 1$.

We need to show $0 < f'(t) = p (t-1)^{p-1} - pt^{p-1}$ and since $p>0$, all we need to show is that $(t-1)^{p-1}>t^{p-1}$, $\forall t > 1$ and $0<p<1$.

Since $0<p<1$, we need to show $t^{1-p} > (t-1)^{1-p}$ which is true since $p<1$ and $t>t-1>0$.

This follows from the more general inequality \begin{equation}(1)\qquad\qquad|x+y|^p\le |x|^p+|y|^p\qquad\qquad\text{(for $x,y\in\mathbb{C}$ and $0\lt p\le1$)}\end{equation} Indeed, if we replace $x$ by $x-y$ in (1) we get $$|x|^p\le |x-y|^p+|y|^p$$ which imply $$|x|^p-|y|^p\le |x-y|^p$$ To prove (1), first note $$|x+y|^p\le(|x|+|y|)^p$$ Hence it is sufficient to prove (1) for $x,y\ge0$ in which case we may apply Sivaram's argument in the previous answer.