It is not true. The function $f(x) = \frac{1}{2} \|x\|^2$ has Hessian $I$ which is positive definite everywhere, but only $x=0$ is a minimum.
If you assume that $a$ is a stationary point of $f$, and that $f$ is twice continuously differentiable at $a$, then the statement is locally true.
Taylor's theorem gives
$f(a+h) = f(a) + \frac{ \partial f(a) }{ \partial x } h + \frac{1}{2} \langle h, \frac{ \partial^2 f(\xi) }{ \partial x^2 } h \rangle$, for some $\xi$ on the line between $a$ and $a+h$.
We have assumed that $a$ is a stationary point, hence $\frac{ \partial f(a) }{ \partial x } = 0$.
Since $\frac{ \partial^2 f(a) }{ \partial x^2 } > 0$, we have $\frac{ \partial^2 f(\xi) }{ \partial x^2 } > 0$ for $\xi$ in some neighborhood $U$ of $a$. Hence for $\xi \in U$, we have $\langle h, \frac{ \partial^2 f(\xi) }{ \partial x^2 } h \rangle \ge 0$. Combining these gives $f(a+h) \ge f(a)$ for all $a+h \in U$. Hence $f(a)$ is a local minimum.
In fact, the above can be tweaked slightly to show that, in fact, $a$ is a strict local minimizer:
Let $\phi(x,h) = \langle h, \frac{ \partial^2 f(x) }{ \partial x^2 } h \rangle$, and $\eta(x) = \min_{\|h\| = 1} \phi(x,h)$. A little work shows that $\eta$ is continuous at $x=a$, and $\eta(a) >0$. Hence for some neighborhood $V$ of $a$, and some $\lambda>0$, we have $\eta(x) \ge \lambda > 0$ for all $x \in V$. It follows from this that $\phi(x,h) \ge \lambda \|h\|^2$ for $x \in V$.
Then, for $a+h \in U \cap V$, we have $f(a+h) \ge f(a) + \lambda \|h\|^2$.
Since the hessian is positive semidefinite for all $x$, the function is convex (though not strictly convex). So the stationary point is a minimum, and a global minimum in fact (by convexity). Think of it this way - the function is increasing in the direction of the eigenvector with eigenvalue 5, and flat in the direction of the eigenvector with eigenvalue 0. If the Hessian were negative semidefinite, you would have a global maximum.
Best Answer
You can even use the univariate Taylor's theorem.
Let $x_0$ satisfy $\nabla L(x_0)=0$. Fix a direction $h \in V$. Define the univariate function $f: \mathbb{R} \to \mathbb{R}$ by $f(t) = L(x_0 + th)$.
Note that $f'(t) = \langle \nabla L(x_0 + th), h\rangle$ and $f''(t) = \langle h, (\nabla^2 L(x_0 + th))h\rangle$. In particular,
Taylor's theorem implies $$f(t) = f(0) + f'(0) t + \frac{f''(\xi_t)}{2} t^2$$ for some $\xi_t$ between $0$ and $t$. Plugging in the definition of $f$ and using the above two facts, we can rewrite the above as $$L(x_0 + th) = L(x_0) + 0 + \frac{1}{2} \langle h, (\nabla^2 L(x_0 + \xi_t h)) h\rangle t^2 \ge L(x_0).$$
Since $h$ was arbitrary, we have $L(x) \ge L(x_0)$ for all $x$.