Incorrect theorem: Suppose $A$ is a real matrix, if $\forall x \neq 0,\ x \in \Bbb R^n,\ x^TAx > 0$, then all eigenvalues $> 0$.
False proof: If $Ax = \lambda x$, then $$ x^TAx = \lambda x^Tx = \lambda ||x||^2> 0$$
Since $||x||^2> 0$, so $\lambda > 0$ q.e.d.
So I'm pretty sure this isn't true, since we can take a rotation matrix as a counter-example. However, I can't find the problem with this false proof.