A direct proof of a property of symmetric 2x2-determinants Let $f(a,b,c)=\det\begin{pmatrix}a &b\\ b& c\end{pmatrix}\in\mathbb{R}[a,b,c]$ be the determinant of a $2 \times 2$ real symmetric matrix. 
Let $f(x_i,y_i,z_i)\geq 0$, $x_i\geq 0$, $z_i\geq 0$ for $i\in\{1,2\}$. 
(Edit: i.e. the matrix is positive semidefinite).
Then
$$
f(x_1+x_2,y_1+y_2,z_1+z_2)\geq 0 \tag{$\ast$}
$$
(something that can be formulated as convexity of the cone of symmetric $2 \times 2$ positive semidefinite (p.s.d.) matrices).
Indeed, $(*)$ can be seen by viewing $f$ as the discriminant of the quadratic polynomial $$F(a,b,c)(t)=at^2+2bt+c \tag{**}$$ with $a\geq 0$, $c\geq 0$, observing that then $f(a,b,c)\geq 0$ iff  $F(a,b,c)(t)\geq 0$ for all $t$, and seeing that $$F(x_1+x_2,y_1+y_2,z_1+z_2)(t)=F(x_1,y_1,z_1)(t)+F(x_2,y_2,z_2)(t)\geq 0$$
The question is whether there is a (more) direct proof of $(*)$, and what kinds of generalisations are known. E.g. I am interested in the situation where $a,b,c$ are multivariate polynomials, and under which conditions $f(a,b,c)$ is a sum of squares (s.o.s.) of polynomials---with potential applications to efficient s.o.s. decompositions.
Edit: As well, is there an analogue of $(**)$ for higher order positive semidefinite matrices?
 A: Here is a direct proof: Simple computations show that
$$f\left(x_1+x_2,y_1+y_2,z_1+z_2\right) = f\left(x_1,y_1,z_1\right) + f\left(x_2,y_2,z_2\right) + \left(x_1z_2+x_2z_1-2y_1y_2\right).$$
It thus remains to check that all three addends on the right hand side are nonnegative. For $f\left(x_1,y_1,z_1\right)$ and $f\left(x_2,y_2,z_2\right)$, this follows straight from the assumptions. For $x_1z_2+x_2z_1-2y_1y_2$, we have to prove that $x_1z_2+x_2z_1 \geq 2y_1y_2$. But thanks to the nonnegativity of $x_i$ and $z_i$, we have $x_1 z_1 \geq y_1^2$ (since $x_1 z_1 - y_1^2 = f\left(x_1, y_1, z_1\right) \geq 0$) and $x_2 z_2 \geq y_2^2$ (similarly), so that the AM-GM inequality yields
$x_1z_2+x_2z_1 \geq 2 \sqrt{x_1z_2 \cdot x_2z_1} = 2 \left(\underbrace{x_1 z_1}_{\geq y_1^2} \underbrace{x_2 z_2}_{\geq y_2^2} \right)^{1/2} \geq 2 \left( y_1^2 y_2^2 \right)^{1/2} = 2 \left| y_1 y_2 \right| \geq 2 y_1 y_2$,
which is precisely what we needed to prove.
Note that this argument might not help you with your question about sum-of-squares decompositions. I am not actually sure what exactly constitutes a sum-of-squares decomposition for an inequality that only holds under assumptions...
