81. By Cauchy-Schwarz inequality,
$$\begin{aligned} 3\left(x^{2}+y^{2}+z^{2}\right) \geqslant & (x+y+z)^{2}=(x+y+z)(x+y+z) \geqslant \\ & 3 \sqrt[3]{x y z}(x+y+z)= \\ & 3(x+y+z) \end{aligned}$$

Therefore,
$$\begin{array}{l} \text { Hence } x^{2}+y^{2}+z^{2} \geqslant x+y+z \\ 2\left[\left(x^{2}+y^{2}+z^{2}\right)-(x y+y z+z x)\right]=(x-y)^{2}+(y-z)^{2}+(z-x)^{2} \geqslant 0 \end{array}$$

By the AM-GM inequality,
$$x^{2}+y^{2}+z^{2} \geqslant 3 \sqrt[3]{(x y z)^{2}}=3$$

Thus,
$$\begin{array}{l} 4\left(x^{2}+y^{2}+z^{2}\right)-[x+y+z+2(x y+y z+z x)+3]= \\ \left(x^{2}+y^{2}+z^{2}\right)-(x+y+z)+2\left[\left(x^{2}+y^{2}+z^{2}\right)-(x y+y z+z x)\right]+ \\ \left(x^{2}+y^{2}+z^{2}\right)-3 \geqslant 0 \end{array}$$

That is,
$$4\left(x^{2}+y^{2}+z^{2}\right) \geqslant x+y+z+2(x y+y z+z x)+3$$

By Cauchy-Schwarz inequality,
$$\begin{array}{l} \frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)}= \\ \frac{x^{4}}{x(1+y)(1+z)}+\frac{y^{4}}{y(1+z)(1+x)}+\frac{z^{4}}{z(1+x)(1+y)} \geqslant \\ \frac{\left(x^{2}+y^{2}+z^{2}\right)^{2}}{x(1+y)(1+z)+y(1+z)(1+x)+z(1+x)(1+y)}= \\ \frac{\left(x^{2}+y^{2}+z^{2}\right)^{2}}{x+y+z+2(x y+y z+z x)+3} \geqslant \frac{\left(x^{2}+y^{2}+z^{2}\right)^{2}}{4\left(x^{2}+y^{2}+z^{2}\right)}= \\ \frac{x^{2}+y^{2}+z^{2}}{4} \geqslant \frac{3 \sqrt[3]{(x y z)^{2}}}{4}=\frac{3}{4} \end{array}$$