4.5.2 ** Find the largest constant $c$ such that for all real numbers $x$, $y$ satisfying $x>0, y>0, x^{2}+y^{2}=1$, we always have $x^{6}+y^{6} \geqslant c x y$.
Substitute $x=y=\frac{\sqrt{2}}{2}$ into the inequality $x^{6}+y^{6} \geqslant c x y$, we know $c \leqslant \frac{x^{6}+y^{6}}{x y}=2 x^{4}=2 \cdot \left(\frac{\sqrt{2}}{2}\right)^{4}=\frac{1}{2}$. Below 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.
(1) $\Leftrightarrow\left(x^{2}+y^{2}\right)\left(x^{4}-x^{2} y^{2}+y^{4}\right) \geqslant \frac{1}{2} x y$
$$
\Leftrightarrow\left(x^{2}+y^{2}\right)^{2}-3 x^{2} y^{2} \geqslant \frac{1}{2} x y \Leftrightarrow 6(x y)^{2}+x y-2 \leqslant 0 .
$$

Let $u=x y$, since $0<x y \leqslant \frac{x^{2}+y^{2}}{2}=\frac{1}{2}$, i.e., $0<u \leqslant \frac{1}{2}$, set $f(u)=6 u^{2}+u-2, 0<u \leqslant \frac{1}{2}$, then $f(u)$ is increasing on $\left(0, \frac{1}{2}\right]$, and $f(0)=-2<0, f\left(\frac{1}{2}\right)=0$. Therefore, on $\left(0, \frac{1}{2}\right]$, $f(u) \leqslant 0$, thus $6(x y)^{2}+x y-2 \leqslant 0$.
In summary, the maximum value of $c$ is $\frac{1}{2}$.
Comment: To find the maximum or minimum value of a parameter contained in an inequality, we can first estimate the upper or lower bound of the parameter, and then prove that the inequality always holds for this bound.

Of course, we can also consider this problem in this way: find the minimum value of $\frac{x^{6}+y^{6}}{x y}$, this minimum value is the maximum value of $c$.
$$
\frac{x^{6}+y^{6}}{x y}=\frac{x^{4}-x^{2} y^{2}+y^{4}}{x y}=\frac{1}{x y}-3 x y .
$$

Since $0<x y \leqslant \frac{1}{2}$, so $\frac{1}{x y} \geqslant 2, -3 x y \geqslant-\frac{3}{2}$. 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 $\frac{1}{2}$, hence the maximum value of $c$ is $\frac{1}{2}$.