Analysis: The left side of the equation to be proven is the product of two trinomials, while the right side is the constant 9. If we use the arithmetic mean inequality once, $a, b, c \in \mathbb{R}^{+} \Rightarrow a+b+c \geqslant 3 \sqrt[3]{a b c}$, a coefficient 3 will appear, so the constant 9 on the right side $=3 \times 3$. Therefore, from $a+b+c \geqslant 3 \sqrt[3]{a b c}>0, \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant 3 \sqrt[3]{\frac{1}{a b c}}>0$, multiplying the two inequalities will suffice.