The classification of symmetric $2\times 2$ real matrices (or bilinear symmetric $2$-forms, or quadratic $2$-forms) through trace and determinant can be obtained in different ways, depending on the machinery one accepts. From more to less:
1) Spectral theorem. Then one knows the classification is done through eigenvalues. For instance positive semidefinite means two positive eigenvalues $\lambda>0,\mu>0$, which is equivalent to $\lambda\cdot\mu>0,\,\lambda+\mu>0$, that is determinant and trace both positive. Honestly, I think that for $2\times 2$ matrices this is too heavy.
2) Canonical forms. Any symmetric $2\times 2$ real matrix $A$ is equivalent to one of the following five canonical forms
$$
\begin{pmatrix}1&0\\0&1\end{pmatrix},\,
\begin{pmatrix}-1&0\\0&-1\end{pmatrix},\,
\begin{pmatrix}1&0\\0&0\end{pmatrix},\,
\begin{pmatrix}-1&0\\0&0\end{pmatrix},\,
\begin{pmatrix}1&0\\0&-1\end{pmatrix}.
$$
The matrix $A$ shares with its canonical form the sign of the determinant (including being $0$). Thus we see that $\det>0$ immediately gives $A$ definite, and it remains to distinguish whether $A$ is positive or negative. In any case, the two entries in the diagonal of $A$ have the same sign, hence the sign of their sum, which is the trace of $A$. Thus $\det(A)>0$, tr$(A)>0$ means positive definite.
3) Nothing. In other words just from the definition. Let $A=
\begin{pmatrix}a&b\\b&c\end{pmatrix}$.
Then the corresponding quadratic form is
$q(x,y)=ax^2+2bxy+cy^2$, and we have to study the sign variations of this function for $(x,y)\ne(0,0)$. For instance
$f(x,0)=ax^2>0$ if and only if $a>0$. Then if $y\ne0$ we can write:
$$
\frac{1}{y^2}q(x,y)=at^2+2bt+c=P(t),\quad t=\frac{x}{y},
$$
and we discuss the signs of $P(t)$. For $t$ big enough, $P(t)>0$, since $a>0$. Then $P(t)>0$ for all $t$ means the polinomial has no zero, that is, its discriminant is negative, which gives
$$
0>\varDelta=b^2-ac=-\det(A).
$$
And we get the condition $\det(A)>0$. Thinking this over one realizes this characterizes being positive semidefinite (that is, is a back and forth argument). And trace? Since $0<\det=ac-b^2$, and $a>0$, necessarily $c>0$ and trace$=a+c>0$.
Best Answer
If $M$ is an $n\times n$ Hermitian and positive definite matrix, it has a Cholesky factorisation: there exists a nonsingular (the result is trivial if $M$ were singular, that is, positive semi-definite) triangular (pick an upper) matrix $U$ such that $M=U^*U$. Now: $$ \det(M)=\det(U^*U)=\det(U^*)\det(U)=\overline{\det(U)}\det(U)=|\det(U)|^2. $$ Now use the Hadamard's inequality which states that an $n\times n$ square matrix $P=[p_1,\ldots,p_n]$ ($p_i$ are the columns of $P$) satisfies $|\det(P)|\leq \prod_{i=1}^n\|p_i\|_2$. Use it on $U=[u_1,\ldots,u_n]$: $$ |\det(U)|\leq\prod_{i=1}^n\|u_i\|_2. $$ Now, what is the diagonal entry $m_{ii}$ of $M$? Well, it is the product $m_{ii}=u_i^*u_i=\|u_i\|_2^2$. So: $$ \det(M)=|\det(U)|^2\leq\prod_{i=1}^n\|u_i\|_2^2=\prod_{i=1}^n m_{ii}. $$