Example 4 Let $\lambda>0$, find the largest constant $c=c(\lambda)$, such that for all non-negative real numbers $x, y$, we have
$$
x^{2}+y^{2}+\lambda x y \geqslant c(x+y)^{2} .
$$
Solution: We consider two cases for $\lambda$:
(i) When $\lambda \geqslant 2$, we have
$$
x^{2}+y^{2}+\lambda x y \geqslant x^{2}+y^{2}+2 x y=(x+y)^{2} \text {. }
$$

The equality holds when $x y=0$.
(ii) When $0<\lambda<2$, we have
$$
\begin{array}{l}
x^{2}+y^{2}+\lambda x y=(x+y)^{2}-(2-\lambda) x y \\
\geqslant(x+y)^{2}-(2-\lambda)\left(\frac{x+y}{2}\right)^{2} \\
=\frac{2+\lambda}{4}(x+y)^{2} .
\end{array}
$$

The equality holds when $x=y$.
Therefore, the largest constant
$$
c(\lambda)=\left\{\begin{array}{ll}
1, & \lambda \geqslant 2, \\
\frac{2+\lambda}{4}, & 0<\lambda<2 .
\end{array}\right.
$$