14. By Cauchy-Schwarz inequality,
$$\begin{array}{l} \left(\frac{x \sqrt{x}}{y+z}+\frac{y \sqrt{y}}{z+x}+\frac{z \sqrt{z}}{x+y}\right)[\sqrt{x}(y+z)+\sqrt{y}(z+x)+\sqrt{z}(x+y)] \geqslant \\ (x+y+z)^{2}=1 \end{array}$$

Assume without loss of generality that $x \geqslant y \geqslant z$, then by Chebyshev's inequality,
$$\begin{array}{l} \sqrt{x}(y+z)+\sqrt{y}(z+x)+\sqrt{z}(x+y) \leqslant \\ \frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})((y+z)+(z+x)+(x+y))}{3}= \\ \frac{2(\sqrt{x}+\sqrt{y}+\sqrt{z})}{3} \end{array}$$

Again by Cauchy-Schwarz inequality,
$$\sqrt{x}+\sqrt{y}+\sqrt{z} \leqslant \sqrt{3(x+y+z)}=\sqrt{3}$$

Therefore,
$$\frac{x \sqrt{x}}{y+z}+\frac{y \sqrt{y}}{z+x}+\frac{z \sqrt{z}}{x+y} \geqslant \frac{\sqrt{3}}{2}$$