4. Find the largest constant $k$, such that for all real numbers $a, b, c, d$ in $[0,1]$, the inequality $a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \geqslant k\left(a^{3}+b^{3}+c^{3}+d^{3}\right)$ holds.
4. 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 $a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \geqslant 2\left(a^{2}+b^{2}+c^{2}+d^{2}\right)$. First, we prove a lemma.
Lemma If $x, y \in[0,1]$, then $x^{2} y+1 \geqslant x^{2}+y^{2}$.
This inequality is equivalent to $(y-1)\left(x^{2}-y-1\right) \geqslant 0$.
Thus, inequality (*) holds. By substituting $a, b$ and $b, c$ and $c, d$ and $d, a$ for $x, y$ in inequality (*), we get
$$
a^{2} b+1 \geqslant a^{2}+b^{2}, b^{2} c+1 \geqslant b^{2}+c^{2}, c^{2} d+1 \geqslant c^{2}+d^{2}, d^{2} a+1 \geqslant d^{2}+a^{2} \text {. }
$$

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

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