46. By Cauchy-Schwarz inequality (square mean is no less than arithmetic mean),
$$\begin{array}{l} \sqrt{2} \sqrt{a+b}=2 \sqrt{\frac{a b}{a+b}} \sqrt{\frac{1}{2}\left(2+\frac{a^{2}+b^{2}}{a b}\right)} \geqslant \\ 2 \sqrt{\frac{a b}{a+b}} \cdot \frac{1}{2}\left(\sqrt{2}+\sqrt{\frac{a^{2}+b^{2}}{a b}}\right)=\sqrt{\frac{2 a b}{a+b}}+\sqrt{\frac{a^{2}+b^{2}}{a+b}} \end{array}$$

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

By Hölder's inequality,
$$\left(\sqrt{\frac{2 a b}{a+b}}+\sqrt{\frac{2 b c}{b+c}}+\sqrt{\frac{2 c a}{c+a}}\right)^{2}\left(\frac{a+b}{2 a b}+\frac{b+c}{2 b c}+\frac{c+a}{2 c a}\right) \geqslant 27$$

Therefore,
$$\sqrt{\frac{2 a b}{a+b}}+\sqrt{\frac{2 b c}{b+c}}+\sqrt{\frac{2 c a}{c+a}} \geqslant 3 \sqrt{\frac{3 a b c}{a b+b c+c a}} \geqslant 3$$