Prove: From the Cauchy-Schwarz inequality and the formula for the area of a triangle,
$$16 S^{2}=\left(a^{2}+b^{2}+c^{2}\right)^{2}-2\left(a^{4}+b^{4}+c^{4}\right),$$

we get
$$\begin{array}{l} 16 S S_{1}+2\left(a^{2} a_{1}^{2}+b^{2} b_{1}^{2}+c^{2} c_{1}^{2}\right) \\ =4 S \cdot 4 S_{1}+\sqrt{2} a^{2} \cdot \sqrt{2} a_{1}^{2}+ \\ \sqrt{2} b^{2} \cdot \sqrt{2} b_{1}^{2}+\sqrt{2} c^{2} \cdot \sqrt{2} c_{1}^{2} \\ <\sqrt{16 S^{2}+2 a^{4}+2 b^{4}+2 c^{4}} \text {. } \\ \sqrt{16 S_{1}^{2}+2 a_{1}^{4}+2 b_{1}^{4}+2 c_{1}^{4}} \\ =\left(a^{2}+b^{2}+c^{2}\right)\left(a_{1}^{2}+b_{1}^{2}+c_{1}^{2}\right) . \end{array}$$

Therefore,
$$\left(a^{2}+b^{2}+c^{2}\right)\left(a_{1}^{2}+b_{1}^{2}+c_{1}^{2}\right)-$$
$$\begin{aligned} & 2\left(a^{2} a_{1}^{2}+b^{2} b_{1}^{2}+c^{2} c_{1}^{2}\right) \\ \geqslant & 16 S S_{1} . \end{aligned}$$

Equality holds if and only if
$$\begin{array}{l} \frac{\sqrt{2} a^{2}}{\sqrt{2} a_{1}^{2}}=\frac{\sqrt{2} b^{2}}{\sqrt{2} b_{1}^{2}}=\frac{\sqrt{2} c^{2}}{\sqrt{2} c_{1}^{2}}=\frac{4 S}{4 S_{1}} \\ \Leftrightarrow \triangle A B C \backsim \triangle A_{1} B_{1} C_{1} . \end{array}$$