Example 6 is the same as Example 5.

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Analyze and Prove: Carefully read the trigonometric substitution method in the original publication. Its core involves rearranging equations and using the AM-GM inequality for two variables, similarly deriving two inequalities, and finally multiplying the three inequalities of the same direction to obtain the proof. In fact, such a way of thinking can directly lead to a solution from the given equation without substitution.

Since \(\frac{a^{2}}{a^{2}+1}+\frac{b^{2}}{b^{2}+1}+\frac{c^{2}}{c^{2}+1}=1\),
we have \(\frac{b^{2}}{b^{2}+1}+\frac{c^{2}}{c^{2}+1}=1-\frac{a^{2}}{a^{2}+1}\) (rearranging),
which means \(\frac{1}{a^{2}+1}=\frac{b^{2}}{b^{2}+1}+\frac{c^{2}}{c^{2}+1}\)
\(\geqslant 2 \sqrt{\frac{b^{2}}{b^{2}+1} \cdot \frac{c^{2}}{c^{2}+1}}\)

Thus, \(\frac{1}{a^{2}+1} \geqslant \frac{2 b c}{\sqrt{\left(b^{2}+1\right)\left(c^{2}+1\right)}} > 0\) (simplifying and transforming).
Similarly, we get \(\frac{1}{b^{2}+1} \geqslant \frac{2 c a}{\sqrt{\left(c^{2}+1\right)\left(a^{2}+1\right)}} > 0\) (symmetric thinking),
\(\frac{1}{c^{2}+1} \geqslant \frac{2 a b}{\sqrt{\left(a^{2}+1\right)\left(b^{2}+1\right)}} > 0\)

Multiplying these three inequalities of the same direction, we simplify to get \(1 \geqslant 8(a b c)^{2}\), so \(a b c \leqslant \frac{\sqrt{2}}{4}\).