Example 16 If $\lambda>0$ for any non-negative real numbers $x_{1}$, $x_{2}$ there is $x_{1}^{2}+x_{2}^{2}+\lambda x_{1} x_{2} \geqslant c\left(x_{1}+x_{2}\right)^{2}$, then the largest constant $c=c(\lambda)=$ $\qquad$ .
Analysis: (1) When $\lambda \geqslant 2$,
$$
\begin{array}{l}
x_{1}^{2}+x_{2}^{2}+\lambda x_{1} x_{2} \\
\geqslant x_{1}^{2}+x_{2}^{2}+2 x_{1} x_{2}=\left(x_{1}+x_{2}\right)^{2},
\end{array}
$$

The equality holds when $x_{1}=x_{2}=0$.
$$
\begin{array}{l}
\text { (2) When } 0<\lambda<2 \text {, } \\
x_{1}^{2}+x_{2}^{2}+\lambda x_{1} x_{2} \\
=\left(x_{1}+x_{2}\right)^{2}-(2-\lambda) x_{1} x_{2} \\
\geqslant\left(x_{1}+x_{2}\right)^{2}-(2-\lambda)\left(\frac{x_{1}+x_{2}}{2}\right)^{2} \\
=\frac{2+\lambda}{4}\left(x_{1}+x_{2}\right)^{2},
\end{array}
$$

The equality holds if and only if $x_{1}=x_{2}$.
From (1) and (2), we have
$$
c(\lambda)=\left\{\begin{array}{cc}
1 & (\lambda \geqslant 2), \\
\frac{2+\lambda}{4} & (0<\lambda<2) .
\end{array}\right.
$$