Assume that $f\_0,f\_1,f\_2$ are polynomial functions of degree two in two variables. This means that the $f\_i$ are linear combinations with real coefficients of $x^2,xy,x,y^2,y,1$.


Consider the function $f = f\_1^2-af\_0f\_2:\mathbb{R}^2\rightarrow\mathbb{R}$ where $a\in \mathbb{R}\_{>0}$. Is it true that for a "random" choice (whit a suitable definition of random) of the $f\_i$ there exists $(x\_0,y\_0)\in\mathbb{R}^2$ such that $f(x\_0,y\_0) \geq 0$.


Clearly, this does not hold for any choice of the $f\_i$. Take for instance $f\_1\equiv 0$, $f\_0 = x^2+1$, $f\_2 = y^2+1$. Then $f(x,y) = -a(x^2y^2+x^2+y^2+1) < 0 $ for all $(x,y)\in\mathbb{R}^2$.


Write
$$f\_0 = a\_1 x^2 + a\_2xy+ a\_3 x+ a\_4 y^2+ a\_5 y+ a\_6;$$
$$f\_1 = b\_1 x^2 + b\_2xy+ b\_3 x+ b\_4 y^2+ b\_5 y+ b\_6;$$
$$f\_2 = c\_1 x^2 + c\_2xy+ c\_3 x+ c\_4 y^2+ c\_5 y+ c\_6;$$
Then $f$ corresponds to the point $(a\_1,\dots,c\_6)\in\mathbb{R}^{18}$.


Taking
$$f\_0 = \epsilon\_0(x^2 + y^2 + 1);$$
$$f\_1 = \epsilon\_1(x^2 + y^2 + 1);$$
$$f\_2 = \epsilon\_2(x^2 + y^2 + 1);$$
we have $f = (\epsilon\_0^2-a\epsilon\_1\epsilon\_2)(x^2 + y^2 + 1)$ which is always negative when $\epsilon\_0^2-a\epsilon\_1\epsilon\_2 < 0$.


But the point $(f\_0,f\_1,f\_2)$ lies inside the linear subspace $a\_2 = a\_3 = a\_5 = b\_2 = b\_3 = b\_5 = c\_2 = c\_3 = c\_5 = 0$.


Thank you very much.



Positivity of real functions in two variables
$\newcommand\R{\mathbb R}\newcommand\c{\mathsf c}\newcommand\ep{\varepsilon}$The answer is no. Indeed, after clarifications given by the OP in comments and in the original post, the question can be stated as follows:



> 
> For $i=0,1,2$, let
> $$f\_i(x,y)=a\_{i,0}+a\_{i,1}x+a\_{i,2}y+a\_{i,3}xy+a\_{i,4}x^2+a\_{i,5}y^2,$$
> where the $a\_{i,j}$ are real numbers.
> For a real $a>0$, let $M\_a$ be the set of all matrices $(a\_{i,j}\colon i=0,1,2,\,j=0,\dots,5)\in\R^{3\times6}$ such that $f(x,y):=f\_1(x,y)^2-af\_0(x,y)f\_2(x,y)\ge0$ for some $(x,y)\in\R^2$.
> Is it true that the dimension (in whatever appropriate sense) of the complement $M\_a^\c$ of $M\_a$ to $\R^{3\times6}$ is strictly less that $18$ (the dimension of $\R^{3\times6}$)?
> 
> 
> 


Let $g(x,y):=1+x^2+y^2$. For real $\ep>0$, let $N\_\ep$ denote the set of all matrices $(a\_{i,j}\colon i=0,1,2,\,j=0,\dots,5)\in\R^{3\times6}$ such that
$|a\_{i,j}-b\_j|<\ep$ for all $i=0,1,2,\,j=0,\dots,5$, where $(b\_0,\dots,b\_5)=(1,0,0,0,1,1)$. Then for $i=0,1,2$ and all real $x,y$ we have
$$|f\_i(x,y)-g(x,y)|<\ep(1+|x|+|y|+|xy|+x^2+y^2)
\le2\ep g(x,y),$$
since $|x|\le(1+x^2)/2$, $|y|\le(1+y^2)/2$, and $|xy|\le(x^2+y^2)/2$.


So, taking now any $a\in(1,2)$ and any $\ep\in(0,\frac{\sqrt a-1}{2(\sqrt a+1)})$, we get ($\ep<1/2$ and) for all real $x,y$
$$f\_1(x,y)^2-af\_0(x,y)f\_2(x,y)\le g(x,y)^2[(1+2\ep)^2-a(1-2\ep)^2]<0.$$
So, $N\_\ep\subseteq M\_a^\c$ and the dimension of $N\_\ep$ is $18$. Thus, the dimension of $M\_a^\c$ is $18$, not $<18$.


