Given $a$, $b$, $x$ are positive integers, and $a \neq b$, $\frac{1}{x}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$. Try to find the minimum value of $x$.

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Given $a$, $b$, $x$ are positive integers, and $a \neq b$, $\frac{1}{x}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$. Try to find the minimum value of $x$.
Solve: It is easy to know that $x=\frac{a^{2} b^{2}}{a^{2}+b^{2}}$.
Let $d=(a, b), a=d a_{0}, b=d b_{0},\left(a_{0}, b_{0}\right)=1$.
Then $x=\frac{d^{2} a_{0}^{2} b_{0}^{2}}{a_{0}^{2}+b_{0}^{2}} \in \mathbf{N}_{+}$.
By $\left(a_{0}^{2}+b_{0}^{2}, a_{0}^{2} b_{0}^{2}\right)=1$, we know $\left(a_{0}^{2}+b_{0}^{2}\right) \mid d^{2}$.
To find the minimum value of $x$, we need $a_{0}, b_{0}, d$ to be as small as possible.
So, $a_{0}=1, b_{0}=2, d=5$.
At this point, $a=5, b=10$.
Therefore, $x_{\text {min }}=20$.