This inequality (Macedonia, 1995) is quite elegant, and a concise proof can be given using the AM-GM inequality. In fact,
$$\begin{aligned} & \sqrt{\frac{x}{y+z}}+\sqrt{\frac{y}{z+x}}+\sqrt{\frac{z}{x+y}} \\ = & \frac{2 x}{2 \sqrt{x(y+z)}}+\frac{2 y}{2 \sqrt{y(z+x)}}+\frac{2 z}{2 \sqrt{z(x+y)}} \\ \geqslant & \frac{2 x}{x+(y+z)}+\frac{2 y}{y+(z+x)}+\frac{2 z}{z+(x+y)}=2 \end{aligned}$$

The equality case, by the problem's condition, is easily excluded.
In this problem, if we let $x=b+c-a, y=c+a-b, z=a+b-c$, we get Problem 548 from the 6th issue of 2001:

Given the side lengths $a, b, c$ of $\triangle ABC$, prove:
$$\begin{array}{l} \sqrt{\frac{b+c-a}{a}}+\sqrt{\frac{c+a-b}{b}}+\sqrt{\frac{a+b-c}{c}} \\ >2 \sqrt{2} \end{array}$$

If the triangle is restricted to being an acute or right triangle, we have a similar problem:

Given the side lengths $a, b, c$ of a non-obtuse $\triangle ABC$, prove:
$$\begin{array}{l} \frac{\sqrt{b^{2}+c^{2}-a^{2}}}{a}+\frac{\sqrt{c^{2}+a^{2}-b^{2}}}{b}+ \\ \frac{\sqrt{a^{2}+b^{2}-c^{2}}}{c} \geqslant 2 \sqrt{2} \end{array}$$