Prove that since $\frac{x+y}{2}+\frac{y+z}{2}+\frac{z+x}{2}=1$, by the generalized Jensen's inequality we have
$$\begin{array}{l} \sum_{\mathrm{cyc}} \frac{x y}{\sqrt{x y+y z}}=\sum_{\mathrm{cyc}} \sqrt{\frac{x^{2} y}{x+z}}=\sum_{\mathrm{cyc}} \frac{x+y}{2} \sqrt{\frac{4 x^{2} y}{(x+y)^{2}(x+z)}} \\ \leqslant \sqrt{\sum_{\mathrm{cyc}} \frac{2 x^{2} y}{(x+y)(x+z)}} . \end{array}$$

It suffices to prove that $\sum_{\mathrm{cyc}} \frac{2 x^{2} y}{(x+y)(x+z)} \leqslant \frac{1}{2}$.
$$\begin{aligned} & \sum_{\mathrm{cyc}} \frac{2 x^{2} y}{(x+y)(x+z)} \leqslant \frac{1}{2} \\ \Leftrightarrow & \sum_{\mathrm{cyc}} \frac{2 x^{2} y}{(x+y)(x+z)} \leqslant \frac{1}{2}(x+y+z) \\ \Leftrightarrow & 4 \sum_{\mathrm{cyc}} x^{2} y(y+z) \leqslant(x+y)(y+z)(z+x)(x+y+z) \\ \Leftrightarrow & 4\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+4 x y z(x+y+z) \leqslant\left[x\left(y^{3}+z^{3}\right)+\right. \\ & \left.y\left(z^{3}+x^{3}\right)+z\left(x^{3}+y^{3}\right)\right]+2\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+4 x y z(x \\ & +y+z) \\ \Leftrightarrow & 2\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right) \leqslant x\left(y^{3}+z^{3}\right)+y\left(z^{3}+x^{3}\right)+z\left(x^{3}+\right. \\ & \left.y^{3}\right)=\left(x^{3} y+x y^{3}\right)+\left(y^{3} z+y z^{3}\right)+\left(x^{3} z+x z^{3}\right) . \end{aligned}$$

By the AM-GM inequality, we have
$$x^{3} y+x y^{3} \geqslant 2 x^{2} y^{2}, y^{3} z+y z^{3} \geqslant 2 y^{2} z^{2}, x^{3} z+x z^{3} \geqslant 2 z^{2} x^{2},$$

Thus, the inequality is clearly true, and the original inequality is proved.