If I am not mistaken more is actually true. Since for $p>1$ the function $f(x)=x^p$ is convex, then by the midpoint inequality one has $f\left(\frac{x+y}{2}\right)\leq \frac{f(x)+f(y)}{2}.$ This translates immediately in $(x+y)^p\leq 2^{p-1}(x^p+y^p),$ which in turn imply your thesis. The case left is then $p=1$ but in this case the result is trivial.
Note that one can say more even in the case $0\leq p<1$. Indeed $\forall t\in [0,+\infty)$ one gets $(1+t)^p\leq 1+t^p.$ To prove the result define $k(t)=(1+t)^p-t^p.$ Then $\forall t\in (0,+\infty)$, we have k'(t)=p[(t+1)^{p-1}-t^{p-1}]<0, and $\lim_{t\to 0}\:k(t)=1,$ therefore the claim follows. Now set $t=\frac{x}{y}$ to conclude $(x+y)^p\leq x^p+y^p.$
EDIT: According to Martin Sleziak's comment, this imply in turn that the function $f(x)=x^p$ is subadditive for $p<1$.
To collect all the informations all at once, one usually says that $\forall p\in [0,+\infty)$ it is true that $(x+y)^p\leq \max(1,2^{p-1})(x^p+y^p).$