Example 6 If positive real numbers $a, b$ and real numbers $x, y$ satisfy $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leqslant 1$,
try to compare $a^{2}+b^{2}$ with $(x+y)^{2}$.
Solve $a^{2}+b^{2} \geqslant\left(a^{2}+b^{2}\right)\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\right)=x^{2}+y^{2}+\frac{b^{2} x^{2}}{a^{2}}+\frac{a^{2} y^{2}}{b^{2}}$
$$
=x^{2}+y^{2}+\left(\frac{b}{a} x-\frac{a}{b} y\right)^{2}+2 x y \geqslant x^{2}+y^{2}+2 x y
$$

Therefore, $a^{2}+b^{2} \geqslant(x+y)^{2}$, equality holds when $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and $\frac{b}{a} x-\frac{a}{b} y=0$,
i.e., when $x=\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}, y=\frac{b^{2}}{\sqrt{a^{2}+b^{2}}}$.
Explanation: Since (1) represents an ellipse and its interior, by comparison, we obtain the elliptical inequality:
(1) If $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leqslant 1$, then $a^{2}+b^{2} \geqslant(x+y)^{2}$.
Special case: If $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, then $a^{2}+b^{2} \geqslant(x+y)^{2}$.
(2) Generalization: By substituting $\frac{x}{r}, \frac{y}{r}$ for $x, y$ in (1), we obtain the following inequality:
If $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leqslant r^{2}$, then $r^{2}\left(a^{2}+b^{2}\right) \geqslant(x+y)^{2}$.