Example 4 Given $a \in(0,1)$, try to determine the range of real number $t$ such that the inequality $a x^{2}+t y^{2} \geqslant(a x+t y)^{2}$ holds for any real numbers $x, y$.

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In this problem, the letter $a$ is a constant, the parameter is $t$, and $x, y$ can be regarded as variables. Since their positions are equal, it is fine to treat $x$ or $y$ as the main variable. For example, if we consider the inequality as an inequality in $x$, we have
$$
a x^{2}+t y^{2} \geqslant a^{2} x^{2}+2 a t x y+t^{2} y^{2} \text {, }
$$

which simplifies to $\left(a^{2}-a\right) x^{2}+2 a t y x+y^{2}\left(t^{2}-t\right) \leqslant 0$.
Since $a \in(0,1)$, we have $a^{2}-a=a(a-1)<0$.
Therefore, the quadratic function $f(x)=\left(a^{2}-a\right) x^{2}+2 a t y x+y^{2}\left(t^{2}-t\right)$ has a downward-opening graph. To ensure that $f(x) \leqslant 0$ for all $x \in \mathbf{R}$, it is necessary that
$$
\begin{aligned}
\Delta & =(2 a t y)^{2}-4\left(a^{2}-a\right) y^{2}\left(t^{2}-t\right) \leqslant 0 \\
& \Rightarrow a^{2} t^{2}-\left(a^{2}-a\right)\left(t^{2}-t\right) \leqslant 0 \\
& \Rightarrow t^{2}-(1-a) t \leqslant 0 \Rightarrow t[t-(1-a)] \leqslant 0,
\end{aligned}
$$

Given $a \in(0,1)$, we get $0 \leqslant t \leqslant 1-a$. Therefore, the range of $t$ that makes the inequality $a x^{2}+t y^{2} \geqslant(a x+t y)^{2}$ hold for all real numbers $x, y$ is $[0,1-a]$.