Since the function $z = t^m$ ($m$ being a constant greater than $0$) is increasing on the open interval $(0, \infty)$, we have
$$(x - y)(x^{n+3} - y^{n+3}) \geq 0 \\ (x - y)(x^{n-1} - y^{n-1}) \geq 0.$$
$$\begin{align*} & \text{Thus } 2\left(\frac{x^{n+1}}{x+y^3} + \frac{y^{n+1}}{x^3+y}\right) - \left(\frac{x^n}{x+y^3} + \frac{y^n}{x^3+y}\right) \\ &= \frac{2x^{n+1} - x^n(x+y)}{x+y^3} + \frac{2y^{n+1} - y^n(x+y)}{x^3+y} \\ & \qquad + \frac{x^n(x-y)}{x+y^3} + \frac{y^n(y-x)}{x^3+y} \\ & \qquad \frac{x-y}{(x+y^3)(x^3+y)} \left[ (x-y)(x^{n+3}-y^{n+3}) + xy(x-y)(x^{n-1}-y^{n-1}) \right] \\ & \ge 0. \end{align*}$$
$$\begin{align*} \text{So}\quad \frac{x^{n+1}}{x+y^3} + \frac{y^{n+1}}{x^3+y} &\ge \frac{1}{2} \left( \frac{x^n}{x+y^3} + \frac{y^n}{x^3+y} \right) \\ &\ge \frac{1}{2^2} \left( \frac{x^{n-1}}{x+y^3} + \frac{y^{n-1}}{x^3+y} \right) \ge \dots \\ &\ge \frac{2^{n-1}}{2} \frac{x^2}{x+y^3} + \frac{y^2}{x^3+y}. \end{align*}$$
则 $\frac{x^n}{x+y^3} + \frac{y^n}{x^3+y} \ge 2^{2-n} \left(\frac{x^2}{x+y^3} + \frac{y^2}{x^3+y}\right)$.

Let $t = xy$, then
$$t \leq \left(\frac{x+y}{2}\right)^2 = \frac{1}{4}, \text{ and } x^4 + y^4 = 1 - 4t + 2t^2, x^5 + y^5 = 1 - 5t + 5t^2.$$

Thus,
$$\begin{aligned} & \frac{x^2}{x+y^3} + \frac{y^2}{x^3+y} \geq \frac{4}{5} \\ \Leftrightarrow & \quad 5[x^2(x^3+y) + y^2(x+y^3)] \geq 4(x+y^3)(x^3+y) \\ \Leftrightarrow & \quad 5(x^5+y^5+t) \geq 4(x^4+y^4+t+t^3) \\ \Leftrightarrow & \quad 5(1-4t+5t^2) \geq 4(1-3t+2t^2+t^3) \\ \Leftrightarrow & \quad 4t^3-17t^2+8t-1 \leq 0 \\ \Leftrightarrow & \quad (4t-1)(t^2-4t+1) \leq 0 \\ \Leftrightarrow & \quad (1-4t)[t^2+(1-4t)] \geq 0. \end{aligned}$$

Since $1-4t \geq 0$, the last inequality holds, and hence,
$$\frac{x^2}{x+y^3} + \frac{y^2}{x^3+y} \geq \frac{4}{5}.$$

Combining the above, the original inequality holds.