Is this inequality true? Clearly if $a,b >0$ and $p \in \mathbb{N}$
$$
a^{p} + b^{p} \le (a+b)^{p}
$$
Is there a constante $C = C(p)$ such that if  $a,b >0$ and $p \in \mathbb{N}$ then
\begin{equation}
a^{p} - b^{p} \le C(p)(a-b)^{p} ?
\end{equation}

It is not possible to find a constant $C(p)$ such that
$$
a^{p} - b^{p} \le C(p)(a-b)^{p}
$$
for all $a,b > 0$ when $p > 1$.  For example, let $a = n+1$ and $b = n$.  Then
$$
a^p-b^p = (n+1)^p - n^p = \sum_{k=0}^{p-1} \binom{p}{k} n^k \to \infty
$$
but $a-b = 1$.
