4.5.4 * Find the largest constant $k$ such that for all real numbers $a, b, c, d$ in $[0,1]$, the inequality
$$
a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \geqslant k\left(a^{2}+b^{2}+c^{2}+d^{2}\right)
$$
holds.
First, estimate the upper bound of $k$. When $a=b=c=d=1$, we have $4 k \leqslant 4+4$, which means $k \leqslant 2$.
Next, we prove that for $a, b, c, d \in[0,1]$, the following inequality always holds:
$$
a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \geqslant 2\left(a^{2}+b^{2}+c^{2}+d^{2}\right).
$$

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

This inequality is equivalent to $(y-1)\left(x^{2}-y-1\right) \geqslant 0$. Since $(y-1)\left(x^{2}-y-1\right) \geqslant 0$ is clearly true, (1) holds. By substituting $a, b ; b, c ; c, d ; d, a$ for $x, y$ in (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 4 inequalities, we obtain
$$
a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \geqslant 2\left(a^{2}+b^{2}+c^{2}+d^{2}\right).
$$

Therefore, the maximum value of $k$ is 2.