Prove upper bound inequality for the dimension of the space of symmetric tensors

Question

I want to check that the dimension of the space of symmetric tensors $N(n,m) := dim(Sym^m(\mathbb{R}^n))$ satisfies $N(n,m) \leq \frac{n^m}{m!}(1+\frac{2m^2}{n})$. Thus I need prove inequality.

If $1\leq m \leq \sqrt{n}$ then $$(n+m)^m \leq n^m(1+2m^2/n)$$