I agree that the eigenvalues has to be positive, but I can't find a connection to the determinants of the principal sub-matrices that I can follow$\dots$
In other words, if $A$ is a symmetrical $n\times n$-matrix that's positive definite, with elements $a_{ii}$. Why must all the principal sub-matrices' determinants be positive (among others $a_{11}*a_{22}-a_{12}*a_{21}>0$)?
Suppose that a symmetric matrix $A=\left(a_{ij}\right)\in {\sf M}_{n\times n}(\mathbb{R})$ is positive definite, then given $1\le k\le n$, we consider its principal $k\times k$ submatrix $A_k$. Since for any non-zero vector ${\bf x}\in\mathbb{R}^n$, we have $${\bf x}^\top A{\bf x}>0.$$ So we let ${\bf x}=(x_1,x_2,\ldots,x_k,0,0\cdots,0)^\top\in\mathbb{R}^n$ be such that the last $n-k$ entries are zero, then we see that \begin{align} {\bf x}^\top{A}{\bf x} &=\begin{pmatrix} x_1&\cdots&x_k&0&\cdots&0 \end{pmatrix}A \begin{pmatrix} x_1\\\vdots\\x_k\\0\\\vdots\\0 \end{pmatrix}\\ &=\begin{pmatrix} \displaystyle\sum_{i=1}^ka_{i1}x_i &\cdots& \displaystyle\sum_{i=1}^ka_{ik}x_i &0&\cdots&0 \end{pmatrix} \begin{pmatrix} x_1\\\vdots\\x_k\\0\\\vdots\\0 \end{pmatrix}\\ &=\sum_{j=1}^k\sum_{i=1}^ka_{ij}x_ix_j, \end{align} which is equal to \begin{align} ({\bf x}^{(k)})^\top{A_k}{\bf x}^{(k)}, \end{align} where ${\bf x}^{(k)}\in\mathbb{R}^k$ such that the entries of ${\bf x}^{(k)}$ are the first $k$ entries of ${\bf x}$. Also, it is clear that $A_k$ is symmetric. So we conclude that $A_k$ is also positive definite, and hence $\det(A_k)>0$.