139. $\frac{x^{2}}{x+y^{2}}=\frac{x\left(x+y^{2}\right)-x y^{2}}{x+y^{2}}=x-\frac{x y^{2}}{x+y^{2}} \geqslant x-\frac{x y^{2}}{2 \sqrt{x y^{2}}}=x-\frac{y \sqrt{x}}{2}$

Similarly, we get
$$\begin{array}{l} \frac{y^{2}}{y+z^{2}} \geqslant y-\frac{z \sqrt{y}}{2} \\ \frac{z^{2}}{z+x^{2}} \geqslant z-\frac{x \sqrt{z}}{2} \end{array}$$

Since $x+y+z=3$, we only need to prove $y \sqrt{x}+z \sqrt{y}+x \sqrt{z} \leqslant 3$. By the AM-GM inequality, we have
$$\begin{array}{r} y \sqrt{x}+z \sqrt{y}+x \sqrt{z} \leqslant y \frac{x+1}{2}+z \frac{y+1}{2}+x \frac{z+1}{2} \leqslant \\ \frac{x+z+y+z x+y}{2} \leqslant \\ 2 \end{array}$$

Or by the Cauchy-Schwarz inequality,
$$\begin{array}{l} (y \sqrt{x}+z \sqrt{y}+x \sqrt{z})^{2} \leqslant(x y+y z+z x)(y+z+x) \\ \frac{1}{(}(x+y+z)^{2}(y+z+x)=\frac{1}{3}(x+y+z)^{3} \\ y \sqrt{x}+z \sqrt{y}+x \sqrt{z} \leqslant 3 \end{array}$$
Thus, we have proved the required inequality.