Prove that from $Q_{2} \geqslant A_{2}$ we get
$$\begin{aligned} & \sqrt{2} \cdot \sqrt{a+b}=2 \sqrt{\frac{a b}{a+b}} \cdot \sqrt{\frac{1}{2}\left(2+\frac{a^{2}+b^{2}}{a b}\right)} \\ \geqslant & 2 \sqrt{\frac{a b}{a+b}} \cdot \frac{1}{2}\left(\sqrt{2}+\sqrt{\frac{a^{2}+b^{2}}{a b}}\right) \\ = & \sqrt{\frac{2 a b}{a+b}}+\sqrt{\frac{a^{2}+b^{2}}{a+b}} \end{aligned}$$

Similarly,
$$\begin{array}{l} \sqrt{2} \cdot \sqrt{b+c} \geqslant \sqrt{\frac{2 b c}{b+c}}+\sqrt{\frac{b^{2}+c^{2}}{b+c}} \\ \sqrt{2} \cdot \sqrt{c+a} \geqslant \sqrt{\frac{2 c a}{c+a}}+\sqrt{\frac{c^{2}+a^{2}}{c+a}} \end{array}$$

From $Q_{3} \geqslant H_{3}$ we get
$$\begin{array}{c} \sqrt{\frac{\left(\sqrt{\frac{a+b}{2 a b}}\right)^{2}+\left(\sqrt{\frac{b+c}{2 b c}}\right)^{2}+\left(\sqrt{\frac{c+a}{2 c a}}\right)^{2}}{3}} \\ \geqslant \frac{3}{\frac{1}{\sqrt{\frac{a+b}{2 a b}}}+\frac{1}{\sqrt{\frac{b+c}{2 b c}}}+\frac{1}{\sqrt{\frac{c+a}{2 c a}}}} \end{array}$$

Therefore,
$$\begin{aligned} & \sqrt{\frac{2 a b}{a+b}}+\sqrt{\frac{2 b c}{b+c}}+\sqrt{\frac{2 c a}{c+a}} \\ \geqslant & 3 \sqrt{\frac{3}{\left(\sqrt{\frac{a+b}{2 a b}}\right)^{2}+\left(\sqrt{\frac{b+c}{2 b c}}\right)^{2}+\left(\sqrt{\frac{c+a}{2 c a}}\right)^{2}}} \\ = & 3 \sqrt{\frac{3 a b c}{a b+b c+c a}} \geqslant 3 . \end{aligned}$$

Thus, the original inequality holds.