I want to prove that the matrix is positive definite using the fact that:
If $A$ is symmetric and $\langle x, Ax \rangle$ > $0$ for a nonzero vector $x$ then $A$ is positive.
So I have the matrix : $A$ =$\begin{bmatrix} 10&-3&0\\-3&10&0\\0&0&9\end{bmatrix}$. And I already checked that $A$ is symmetric, as $A=A^T$
and then I also compute that
$\langle x, Ax \rangle = 10a^2+10b^2+9c^2 -6ab$, so is it that true that eventhough I have one negative number (-6ab), $\langle x,Ax \rangle$ is still positive ?
Having negative terms in the resulting polynomial does not necessarily mean that the overall polynomial function attains values $\leq 0$. Indeed, we can prove it to be strictly positive for every nonzero vector $x = (a,b,c)^T$.
So in your last step you reached
$$\langle x, Ax \rangle = 10a^2 + 10b^2 + 9c^2 - 6ab$$
Do you notice this suspicious mixed term $-6ab$? That should ring a bell! It commonly occurs in the identity $(u-v)^2 = u^2 - 2uv + v^2$. Let's see if we can manipulate it to reach the LHS of the identity:
$$ \begin{align*} &10a^2 + 10b^2 + 9c^2 - 6ab\\ =& ??? + ??? + 9c^2 + 3(a^2-2ab+b^2)\\ =& 7a^2 + 7b^2 + 9c^2 + 3(a^2-2ab+b^2)\\ =& 7a^2 + 7b^2 + 9c^2 + 3(a-b)^2 \end{align*} $$ Usually, I would do the step with $???$, but I explicitly listed it for you here. In this step you just attempt to write $-6ab$ in the form of the identity I mentioned and then try to adjust the other summands, so that we indeed have a valid equality from the previous line.
Can you now prove how the expression on the last line is always $> 0$ for every $x = (a, b, c)^T \neq 0$?
By the way, I think in general going the eigenvalue way, i.e. a symmetric matrix is positive definite iff. all eigenvalues are $> 0$, is more systematic and easier.