Example 1 Find the largest constant $c$, such that for real numbers $x, y$ satisfying $x>0, y>0, x^{2}+y^{2}=1$, it always holds that $x^{6}+y^{6} \geqslant c x y$.


When $x=y=\frac{\sqrt{2}}{2}$, the inequality also holds, then,
$$
c \leqslant \frac{x^{6}+y^{6}}{x y}=\frac{1}{2} \text {. }
$$

Next, we prove that under the conditions $x>0, y>0, x^{2}+y^{2}=1$, the inequality
$$
x^{6}+y^{6} \geqslant \frac{1}{2} x y
$$

always holds.
Equation (1) is equivalent to
$$
\left(x^{2}+y^{2}\right)\left[\left(x^{2}+y^{2}\right)^{2}-3 x^{2} y^{2}\right] \geqslant \frac{1}{2} x y,
$$

which simplifies to $6(x y)^{2}+x y-2 \leqslant 0$.
Let $\mu=x y$, since $0<x y \leqslant \frac{x^{2}+y^{2}}{2}=\frac{1}{2}$, we have $0<\mu \leqslant \frac{1}{2}$.
Let $f(\mu)=6 \mu^{2}+\mu-2,0<\mu \leqslant \frac{1}{2}$, then $f(\mu)$ is increasing on $\left(0, \frac{1}{2}\right]$, and $f\left(\frac{1}{2}\right)=0$. Therefore, on $\left(0, \frac{1}{2}\right]$, we have $f(\mu) \leqslant 0$. Thus,
$$
6(x y)^{2}+x y-2 \leqslant 0 \text {. }
$$

In summary, the maximum value of $c$ is $\frac{1}{2}$.
Note: To find the maximum or minimum value of a parameter in an inequality, one can first estimate the upper or lower bound of the parameter, and then prove that the inequality always holds for this upper or lower bound.

This problem can also be considered as follows: Find the minimum value of $\frac{x^{6}+y^{6}}{x y}$, which is the maximum value of $c$. That is,
$$
\begin{aligned}
\frac{x^{6}+y^{6}}{x y} & =\frac{\left(x^{2}+y^{2}\right)\left[\left(x^{2}+y^{2}\right)^{2}-3 x^{2} y^{2}\right]}{x y} \\
& =\frac{1}{x y}-3 x y .
\end{aligned}
$$

Since $0<x y \leqslant \frac{1}{2}$, we have $\frac{1}{x y} \geqslant 2,-3 x y \geqslant-\frac{3}{2}$. Therefore, $\frac{x^{6}+y^{6}}{x y} \geqslant 2-\frac{3}{2}=\frac{1}{2}$.

When $x=y=\frac{\sqrt{2}}{2}$, the equality holds, thus the minimum value of $\frac{x^{6}+y^{6}}{x y}$ is $\frac{1}{2}$, and the maximum value of $c$ is $\frac{1}{2}$.