A thermodynamic inequality (from minimum work)

[mSSM]

EDIT: Turns out, the solution to my question is related to the determinant of a positive definite quadratic form.

This is more or less straight from Landau's Statistical Physics Part 1 (3rd edition), Chapter 21.

I don't understand how the inequality/condition (the last equation in this post) arises. I think it's a mathematical problem, but I wonder if I have ignored some physical reasoning.

The minimum work, which must be done to bring a body in an external medium from its equilibrium state to any neighbouring state:
$$\delta E - T_0 \delta S + P_0 \delta V > 0$$
where $T_0$ and $P_0$ are (constant) temperature and pressure of the external medium, and where for the temperature and the pressure of the body in equilibrium with the external medium we would have: $T=T_0$, and $P=P_0$

Regarding the energy as $E=E(S,V)$, and expanding $\delta E$ as a series:
$$\delta E = \frac{\partial E}{\partial S}\delta S + \frac{\partial E}{\partial V}\delta V + \frac{1}{2} \left[ \frac{\partial^2 E}{\partial S^2} (\delta S)^2+ 2 \frac{\partial^2 E}{\partial S\partial V} \delta S\delta V + \frac{\partial^2 E}{\partial V^2} (\delta V)^2 \right]$$

Note that $\frac{\partial E}{\partial S}=T$ and $\frac{\partial E}{\partial V}=P$. Inserting this into the first equation gives the inequality:
$$\frac{\partial^2 E}{\partial S^2} (\delta S)^2 + 2 \frac{\partial^2 E}{\partial S\partial V} \delta S\delta V+ \frac{\partial^2 E}{\partial V^2} (\delta V)^2> 0$$

If we want this inequality to be true for any values of $\delta S$ and $\delta V$, the following conditions have to hold:
$$\frac{\partial^2 E}{\partial S^2} > 0$$

$$\frac{\partial^2 E}{\partial S^2}\frac{\partial^2 E}{\partial V^2} - \left( \frac{\partial^2 E}{\partial S\partial V} \right)^2 > 0$$

Where does the last condition come from?

 

[eljose79]

The last condition comes from the fact that the quadratic form \frac{\partial^2 E}{\partial S^2}(\delta S)^2+2\frac{\partial^2 E}{\partial S\partial V}\delta S\delta V+\frac{\partial^2 E}{\partial V^2}(\delta V)^2 represents the Hessian matrix of the energy function, which is a positive definite matrix. This means that all of its eigenvalues are positive, and in order for this to be true, the determinant of the matrix must also be positive. Thus, the condition \frac{\partial^2 E}{\partial S^2}\frac{\partial^2 E}{\partial V^2}-\left(\frac{\partial^2 E}{\partial S\partial V}\right)^2>0 is necessary for the inequality to hold for any values of \delta S and \delta V.