We know that if $\nabla f(x_0) = 0$ and $\nabla^2f(x_0) \succ0$ (positive definite) then $x_0$ is the local minima.
So, prove that if $\nabla f(x_0) = 0$ and $\nabla^2f(x_0) \succeq 0$ then $x_0$ couldn't be the local minimum. If possible, don't use counterexamples, but really a proof.
I'm going to give you some examples. You can decide if they answer your proof or not.
Take $f(x, y) = x^2 + y^2$. Then $\nabla f(0, 0) = (0, 0)$, and $$\nabla^2 f(0, 0)=\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \succeq 0.$$ The fact that the matrix is positive definite means that it's also positive semi-definite. Note that $(0, 0)$ is a local (and in fact, global) minimum.
Take $g(x, y) = x^2$. Then $\nabla g(0, 0) = (0, 0)$, and $$\nabla^2 g(0, 0)=\begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix} \succeq 0.$$ Note that $0$ is the minimal value of $g$, making $(0, 0)$ a local (and global) minimum. However, this minimum is achieved all along the line $x = 0$, e.g. at $(0, 1)$ as well.
Take $h(x, y) = x^2 + y^4$. Then $\nabla h (0, 0) = (0, 0)$, and $$\nabla^2 h(0, 0)=\begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix} \succeq 0.$$ Again, we have a global minimum at $(x, y) = (0, 0)$, and this time it's unique. However, the Hessian was only positive semi-definite.
Finally, take $k(x, y) = x^2 - y^4$. Then $\nabla k(0, 0) = (0, 0)$, and $$\nabla^2 k(0, 0)=\begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix} \succeq 0.$$ This time, we have positive semi-definiteness in the Hessian, but no local minimum.