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

By Cauchy-Schwarz inequality, we have $3\left(x^{2}+y^{2}+z^{2}\right) \geqslant(x+y+z)^{2}$, and by the AM-GM inequality,
$$\begin{array}{c} x^{2}+y^{2}+z^{2} \geqslant x y+y z+z x \\ x+y+z \geqslant 3 \sqrt[3]{x y z}=3 \end{array}$$

Therefore,
$$\begin{aligned} 3\left(x^{2}+y^{2}+z^{2}\right) \geqslant & (x+y+z)^{2}=(x+y+z)(x+y+z) \geqslant \\ & 3(x+y+z) \end{aligned}$$

That is,
$$x^{2}+y^{2}+z^{2} \geqslant x+y+z$$

So,
$$\begin{array}{l} x(1+y)(1+z)+y(1+z)(1+x)+z(1+x)(1+y)= \\ (x+y+z)+2(x y+y z+z x)+3 x y z= \\ 3+(x+y+z)+2(x y+y z+z x) \end{array}$$

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

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