5. In the Cartesian coordinate system $x O y$, the area of the figure bounded by the inequalities
$$
y^{100}+\frac{1}{y^{100}} \leqslant x^{100}+\frac{1}{x^{100}}, \quad x^{2}+y^{2} \leqslant 100
$$

is equal to
5. $50 \pi$.

We examine a region that is symmetric with respect to the coordinate axes, so we only need to consider the part where $x>0, y>0$. Let $u=x^{100}$, $v=y^{100}$, then
$$
u+\frac{1}{u} \geqslant v+\frac{1}{v}
$$

After rearrangement, this inequality is equivalent to
$$
(u-v)(u v-1) \geqslant 0 .
$$

Therefore, the solution set of the inequality has the form as shown in the shaded area in Figure 2. Since the shaded part of the circle $x^{2}+y^{2}=100$ is equal in area to the unshaded part, the area of the shaded part is $\frac{1}{2} \times 10^{2} \pi=50 \pi$.