For non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{1}+x_{2}+\cdots+x_{n}=1$. Find the maximum value of $\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)$.
Solution: Use the adjustment method to explore the maximum value of the sum in the problem.
(i) First, for $x, y > 0$, we compare $(x+y)^{4} - (x+y)^{5} + 0^{4} - 0^{5}$ with $x^{4} - x^{5} + y^{4} - y^{5}$:
$$
\begin{array}{l}
(x+y)^{4} - (x+y)^{5} + 0^{4} - 0^{5} - \left(x^{4} - x^{5} + y^{4} - y^{5}\right) \\
= x y \left(4 x^{2} + 6 x y + 4 y^{2}\right) \\
- x y \left(5 x^{3} + 10 x^{2} y + 10 x y^{2} + 5 y^{3}\right) \\
\geqslant \frac{7}{2} x y \left(x^{2} + 2 x y + y^{2}\right) \\
- 5 x y \left(x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3}\right) \\
= \frac{1}{2} x y (x+y)^{2} [7 - 10(x+y)].
\end{array}
$$

As long as $x, y > 0, x + y > 0$, then, among them, there must be two whose sum $\leqslant \frac{2}{3}0, x + y = 1$.
$$

For this case,
$$
\begin{array}{l}
x^{4} - x^{5} + y^{4} - y^{5} \\
= x^{4}(1 - x) + y^{4}(1 - y) \\
= x y \left(x^{3} + y^{3}\right) \\
= x y \left[(x + y)^{3} - 3 x y (x + y)\right] \\
= x y (1 - 3 x y) \\
= \frac{1}{3} (3 x y) (1 - 3 x y).
\end{array}
$$

When $3 x y = \frac{1}{2}$, the above expression reaches its maximum value:
$$
x^{4} - x^{5} + y^{4} - y^{5} = \frac{1}{6} \left(1 - \frac{1}{2}\right) = \frac{1}{12}.
$$

This is the maximum value sought in the problem. The $x_{1}, x_{2}, \cdots, x_{n}$ that can achieve this maximum value have only two that are not equal to 0. Let $x$ and $y$ represent these two numbers, then
$$
x, y > 0, x + y = 1, x y = \frac{1}{6}.
$$

Solving the quadratic equation $\lambda^{2} - \lambda + \frac{1}{6} = 0$ yields
$$
x = \frac{3 + \sqrt{3}}{6}, y = \frac{3 - \sqrt{3}}{6},
$$

or it can also be $x = \frac{3 - \sqrt{3}}{6}, y = \frac{3 + \sqrt{3}}{6}$.
Verification shows that if $x_{1}, x_{2}, \cdots, x_{n}$ have only these two non-zero numbers, then the sum in the problem indeed reaches the maximum value $\frac{1}{12}$.