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](https://arxiv.org/abs/1511.03473).


**Edit:** As well, is there an analogue of $(\*\*)$ for higher order positive semidefinite matrices?



A direct proof of a property of symmetric 2x2-determinants
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...


