Example 13.5 (Huang Chendi) Given $a \geqslant b \geqslant c \geqslant d, a+b+c+d=1$, and $k$ is a given positive real number, find the maximum value of $\lambda$ such that
$$\frac{a-b}{k+c+d}+\frac{b-c}{k+d+a}+\frac{c-d}{k+a+b}+\frac{d-a}{k+b+c} \geqslant \lambda(a-b)(b-c)(c-d)$$
Solving this problem is not easy. First, we observe the original problem and try to guess the maximum value of $\boldsymbol{\lambda}$, which is quite difficult. Notice that both sides of the inequality have $a-b, b-c, c-d, d-a$, so we consider a direct difference substitution. We start by bounding the left side of the inequality to approach the right side, trying to maintain the same conditions for equality in each step of the bounding process to determine the range of $\lambda$. We then use the conditions for equality to show that there exist corresponding $a, b, c, d$. The right side of the inequality is in a product form, suggesting that we might use the AM-GM inequality.

On one hand, let $x=a-b, y=b-c, z=c-d$, thus we have
$$\begin{array}{l}
\frac{a-b}{k+c+d}+\frac{b-c}{k+d+a}+\frac{c-d}{k+a+b}+\frac{d-a}{k+b+c}= \\
\left(\frac{a-b}{k+c+d}+\frac{c-d}{k+a+b}\right)+\left(\frac{b-c}{k+a+d}+\frac{d-a}{k+b+c}\right)= \\
\frac{k(a-b+c-d)+\left(a^{2}-b^{2}+c^{2}-d^{2}\right)}{(k+a+b)(k+c+d)}+ \\
\frac{k(b-c+d-a)+\left(b^{2}-c^{2}+d^{2}-a^{2}\right)}{(k+b+c)(k+d+a)}= \\
\left(k(a-b+c-d)+\left(a^{2}-b^{2}+c^{2}-d^{2}\right)\right) \cdot
\end{array}$$
$$\begin{array}{l}
\left(\frac{1}{(k+a+b)(k+c+d)}-\frac{1}{(k+b+c)(k+a+d)}\right)= \\
\frac{\left(k(a-b+c-d)+\left(a^{2}-b^{2}+c^{2}-d^{2}\right)((k+b+c)(k+d+a)-(k+a+b)(k+c+d))\right.}{(k+a+b)(k+b+c)(k+c+d)(k+d+a)}= \\
\frac{((k+a+b) x+(k+c+d) z)(x+y)(y+z)}{(k+a+b)(k+b+c)(k+c+d)(k+d+a)} \geqslant \\
\frac{8 x y z \sqrt{(k+a+b)(k+c+d)}}{(k+a+b)(k+b+c)(k+c+d)(k+d+a)}= \\
\frac{8 x y z}{\sqrt{(k+a+b)(k+c+d)(k+b+c)^{2}(k+d+a)^{2}}} \geqslant
\end{array}$$
$\frac{64}{(2 k+1)^{3}} x y z$
Thus, we obtain
$$\lambda_{\max } \geqslant \frac{64}{(2 k+1)^{3}}$$