Starting with the terms of the right-hand side, the quadratic-arithmetic-mean inequality yields
$$\begin{aligned} \sqrt{2} \sqrt{a+b} & =2 \sqrt{\frac{a b}{a+b}} \sqrt{\frac{1}{2}\left(2+\frac{a^{2}+b^{2}}{a b}\right)} \\ & \geq 2 \sqrt{\frac{a b}{a+b}} \cdot \frac{1}{2}\left(\sqrt{2}+\sqrt{\frac{a^{2}+b^{2}}{a b}}\right)=\sqrt{\frac{2 a b}{a+b}}+\sqrt{\frac{a^{2}+b^{2}}{a+b}} \end{aligned}$$
and, analogously,
$$\sqrt{2} \sqrt{b+c} \geq \sqrt{\frac{2 b c}{b+c}}+\sqrt{\frac{b^{2}+c^{2}}{b+c}}, \quad \sqrt{2} \sqrt{c+a} \geq \sqrt{\frac{2 c a}{c+a}}+\sqrt{\frac{c^{2}+a^{2}}{c+a}} .$$
Applying the inequality between the arithmetic mean and the squared harmonic mean will finish the proof:
$$\sqrt{\frac{2 a b}{a+b}}+\sqrt{\frac{2 b c}{b+c}}+\sqrt{\frac{2 c a}{c+a}} \geq 3 \cdot \sqrt{\frac{3}{\sqrt{\frac{a+b}{2 a b}}^{2}+\sqrt{\frac{b+c}{2 b c}}^{2}+\sqrt{\frac{c+a}{2 c a}}}}{ }^{2}=3 \cdot \sqrt{\frac{3 a b c}{a b+b c+c a}} \geq 3 .$$