Prove: By the 2-variable mean inequality, we get
$$\frac{a^{2}+b^{2}}{2 a b}+\frac{2 \sqrt{a b}}{a+b} \geq 2 \sqrt{\frac{a^{2}+b^{2}}{2 a b} \cdot \frac{2 \sqrt{a b}}{a+b}}=2 \sqrt{\frac{a^{2}+b^{2}}{\sqrt{a^{3} b}+\sqrt{a b^{3}}}}$$

Therefore, it suffices to prove $\frac{a^{2}+b^{2}}{\sqrt{a^{3} b}+\sqrt{a b^{3}}} \geq 1$, which is equivalent to
$$a^{2}+b^{2} \geq \sqrt{a^{3} b}+\sqrt{a b^{3}}$$

By the 4-variable mean inequality, we get
$$\begin{aligned} a^{2}+b^{2} & =\frac{a^{2}+a^{2}+a^{2}+b^{2}}{4}+\frac{a^{2}+b^{2}+b^{2}+b^{2}}{4} \\ & \geq \sqrt[4]{a^{2} \cdot a^{2} \cdot a^{2} \cdot b^{2}}+\sqrt[4]{a^{2} \cdot b^{2} \cdot b^{2} \cdot b^{2}} \\ & =\sqrt{a^{3} b}+\sqrt{a b^{3}}, \text { Q.E.D. } \end{aligned}$$