Solution: We choose a positive number $l<5$, and apply the AM-GM inequality, we have the following inequalities:
$$\begin{array}{l} l x^{2}+2 y^{2} \geq 2 \sqrt{2 l} x y \\ 2 y^{2}+l z^{2} \geq 2 \sqrt{2} y z \\ (5-l) z^{2}+\frac{1}{2} t^{2} \geq \sqrt{2(5-l)} z t \\ \frac{1}{2} t^{2}+(5-l) x^{2} \geq \sqrt{2(5-l)} t x \end{array}$$

Adding the above inequalities, we get
$$5 x^{2}+4 y^{2}+5 z^{2}+t^{2} \geq 2 \sqrt{2 l}(x y+t z)+\sqrt{2(5-l)}(z t+t x)$$

The condition $x y+y z+z t+t x=1$ suggests we choose a number $l(0 \leq l \leq 5)$, such that $2 \sqrt{2 l}=\sqrt{2(5-l)}$. Simple calculation yields $l=1$, so the minimum value of the expression $5 x^{2}+4 y^{2}+5 z^{2}+t^{2}$ is $2 \sqrt{2}$.

Note: The following general problem can be solved using the same method.

Assume $x, y, z, t$ are arbitrary real numbers. Prove that
$$x^{2}+k y^{2}+z^{2}+l t^{2} \geq \sqrt{\frac{2 k l}{k+l}}(x y+y z+z x+t x)$$