Example 3 Find the largest constant $k$, such that for all real numbers $a, b, c, d$ in $[0,1]$, the inequality
$$
\begin{array}{l}
a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \\
\geqslant k\left(a^{2}+b^{2}+c^{2}+d^{2}\right) .
\end{array}
$$
holds.
Solution: First, estimate the upper bound of $k$.
When $a=b=c=d=1$, we have $4 k \leqslant 4+4, k \leqslant 2$.
Next, we prove that for $a, b, c, d \in[0,1]$, it always holds that
$$
\begin{array}{l}
a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \\
\geqslant 2\left(a^{2}+b^{2}+c^{2}+d^{2}\right) .
\end{array}
$$

First, we prove a lemma.
Lemma If $x, y \in[0,1]$, then
$$
x^{2} y+1 \geqslant x^{2}+y^{2} \text {. }
$$

This inequality is equivalent to $(y-1)\left(x^{2}-y-1\right) \geqslant 0$.
Thus, inequality (2) holds. By substituting $a, b$ and $b, c$ and $c, d$ and $d, a$ for $x, y$ in inequality (1), we get
$$
\begin{array}{l}
a^{2} b+1 \geqslant a^{2}+b^{2}, b^{2} c+1 \geqslant b^{2}+c^{2}, \\
c^{2} d+1 \geqslant c^{2}+d^{2}, d^{2} a+1 \geqslant d^{2}+a^{2} .
\end{array}
$$

Adding the above four inequalities, we obtain
$$
\begin{array}{l}
a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \\
\geqslant 2\left(a^{2}+b^{2}+c^{2}+d^{2}\right) .
\end{array}
$$

In conclusion, the maximum value of $k$ is 2.