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.