Problem says:
Show that if $f:A\subset \mathbb R^2 \rightarrow \mathbb R$ has a critical point $x_0 \in A$ and we let
$\Delta =$determinant of Hessian of $f$
be evaluated at $x_0$, then $\Delta >0$ and $\partial^2 f/\partial x_1 >0$ and $ \partial^2 f/\partial x_1 \partial x_1<0$ imply $f$ has a local maximum at $x_0$
I proved the case when $f$ has a local minimum at $x_0$.
It suffices to show that $\left(\begin{array}{cc} \frac{\partial^{2}f}{\partial x_{1}\partial x_{1}} & \frac{\partial^{2}f}{\partial x_{1}\partial x_{2}}\\ \frac{\partial^{2}f}{\partial x_{2}\partial x_{1}} & \frac{\partial^{2}f}{\partial x_{2}\partial x_{2}} \end{array}\right)$ is positive definite, that is, for any $v=(x,y)^{T}\in\mathbb{R}^{3}\backslash\{(0,0)\} $ ,
$(\begin{array}{cc} x & y\end{array})\left(\begin{array}{cc} a & b\\ b & c \end{array}\right)\left(\begin{array}{c} x\\ y \end{array}\right)>0$
So, $a(x+\frac{b}{a}y)^{2}+(c-\frac{b^{2}}{a})y^{2}>0$
By assumption $a>0$ and $ac-b^{2}>0$ , we get desired one.
But the problem is to show that $\left(\begin{array}{cc} \frac{\partial^{2}f}{\partial x_{1}\partial x_{1}} & \frac{\partial^{2}f}{\partial x_{1}\partial x_{2}}\\ \frac{\partial^{2}f}{\partial x_{2}\partial x_{1}} & \frac{\partial^{2}f}{\partial x_{2}\partial x_{2}} \end{array}\right)$ is negative definite so that f has a local minimum at $x_{0}$ .
$ \left(\begin{array}{cc} \frac{\partial^{2}f}{\partial x_{1}\partial x_{1}} & \frac{\partial^{2}f}{\partial x_{1}\partial x_{2}}\\ \frac{\partial^{2}f}{\partial x_{2}\partial x_{1}} & \frac{\partial^{2}f}{\partial x_{2}\partial x_{2}} \end{array}\right)$ is negative definite is equivalent to $a(x+\frac{b}{a}y)^{2}+(c-\frac{b^{2}}{a})y^{2}<0$ for $(x,y)\in\mathbb{R}^{2}\backslash\{(0,0)\}$ . But given condition says only $a>0$ and $ac-b^{2}<0$ , i.e., $a(x+\frac{b}{a}y)^{2}>0$ and $(c-\frac{b^{2}}{a})y^{2}>0$ . How could I show that the sum of them is negative?