Example 1. For the inequality
$$
x^{2}+4 x y+4 y^{2}+10 x+a y+b>0
$$

to hold for all real numbers $x, y$, determine the conditions that the constants $a, b$ must satisfy.
To make the inequality always hold, we need:
$$
x^{2}+4 x y+4 y^{2}+10 x+a y+b
$$

to be a real quadratic form plus a small positive number $\varepsilon$. We can let
$$
\begin{aligned}
& x^{2}+4 x y+4 y^{2}+10 x+a y+b \\
= & (x+2 y+m)^{2}+\varepsilon, \quad(m \in R)
\end{aligned}
$$

which means
$$
\begin{aligned}
& x^{2}+4 x y+4 y^{2}+10 x+a y+b \\
= & x^{2}+4 x y+4 y^{2}+2 m x+4 m y+m^{2}+\varepsilon
\end{aligned}
$$

According to the conditions for the identity of polynomials, we have
$$
\left\{\begin{array} { l } 
{ 1 0 = 2 m , } \\
{ a = 4 m , } \\
{ b = m ^ { 2 } + \varepsilon }
\end{array} \quad \text { Solving, we get } \left\{\begin{array}{l}
a=20, \\
b=25+\varepsilon
\end{array}\right.\right.
$$

Therefore, when $a=20$ and $b>25$, the inequality
$$
x^{2}+4 x y+4 y^{2}+10 x+a y+b>0
$$

always holds.