Binary quadratic form Let $q(x,y)=ax^2+bxy+cy^2$ is a binary quadratic form. And $q(x,y)>0$ for any $(x,y)\neq(0,0)$. How to prove that $\exists C>0$ such that $q(x,y)>C(x^2+y^2)$
 A: $q$ is continuous and non-zero on the compact set 
$\{ (x, y) : x^2 + y^2 = 1 \}$, therefore
$$
 m :=  \min \{ q(x, y) : x^2 + y^2  = 1 \}
$$
is strictly positive.
Now for arbitrary $(x, y) \ne (0, 0)$,
$$
 (u, v) := (\frac{x}{\sqrt{x^2 + y^2}}, \frac{y}{\sqrt{x^2 + y^2}})
$$
satisfies $u^2 + v^2 = 1$, and therefore
$$
 \frac{q(x, y)}{x^2 + y^2} = q( \frac{x}{\sqrt{x^2 + y^2}}, \frac{y}{\sqrt{x^2 + y^2}}) = q(u, v) \ge m \, .
$$
It follows that for any $0 < C < m$,
$$
 q(x, y) >  C (x^2 + y^2) \quad \text{ for } (x, y) \ne (0, 0) \, .
$$
Of course the strict inequality cannot hold for $(x, y) = (0, 0)$.
A: Recall that a binary quadratic form $ax^2 + bxy + cy^2$ is positive definite if and only if all leading principal minors of the matrix $A = \begin{bmatrix} a & \frac{b}{2} \\ \frac{b}{2} & c \end{bmatrix}$ are positive. In this case this means that $a > 0$ and $\det(A) > 0$.  
Now rephrasing your question, you are asking whether there is a $d > 0$ such that $(a - d)x^2 + bxy + (c - d)y^2$ is still a positive definite quadratic form. 
But since $\det(A) > 0$ the function $d \mapsto \det(A - d \cdot Id)$ (which is continuous in $d$, as it is in fact a polynomial) is positive in some interval around zero. Choose $0 < d' < a$ in said interval. Then $A - d' \cdot Id$ is a positive definite matrix, and you have found a $d' > 0$ such that $q(x, y) > d'(x^2 + y^2)$.
