Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ be a three times differentiable function and its gradient be Lipschitz smooth with constant $L$, or to say, the Hessian of $f$ is Lipschitz continous, i.e. , $$\|\nabla^2f(y) - \nabla^2f(x)\| \le L \|y-x\|,$$ where $x,y \in \mathbb{R}^n$ and $\| \cdot \|$ denotes the Euclidean Norm. My question is whether this inequality is equivalent to the following one, $$\| \nabla f(y) - \nabla f(x) - \nabla^2 f(x) \cdot (y-x) \| \le \frac{L}{2} \| y-x \|^2\quad ?$$ If true, How can I prove this? I only know how to prove the similar expression when $f$ is $L$-smooth, but the dimension of $\nabla f$ is $n$, not $1$, and I am confused.
Any hints or suggestions would be appreciated!!!