22. First, we have
$$\begin{array}{c} \left(\sum_{\mathrm{cyc}} x\right)\left(\sum_{\mathrm{cyc}} \frac{1}{x}\right)-9=\sum_{\mathrm{cyc}} \frac{(x-y)^{2}}{x y} \\ 6\left(\sum_{\mathrm{cyc}} \frac{x}{y+z}\right)-9=\sum_{\mathrm{cyc}} \frac{3(x-y)^{2}}{(x+z)(y+z)} \end{array}$$

From equations (7) and (8), to prove the original inequality, we only need to prove
$$\sum_{\mathrm{cyc}} \frac{(x-y)^{2}}{x y} \geqslant \sum_{\mathrm{cyc}} \frac{3(x-y)^{2}}{(x+z)(y+z)}$$

Let
$$\begin{array}{l} S_{x}=\frac{1}{y z}-\frac{3}{(y+x)(z+x)} \\ S_{y}=\frac{1}{z x}-\frac{3}{(z+y)(x+y)} \\ S_{z}=\frac{1}{x y}-\frac{3}{(x+z)(y+z)} \end{array}$$

Assume without loss of generality that $x \geqslant y \geqslant z$. Clearly, $S_{x} \geqslant 0$. Since $y+z \geqslant x$, it is easy to see that $x y+z y+ y^{2}-2 x z \geqslant 0$, which means $S_{y} \geqslant 0$. If $S_{z} \geqslant 0$, the inequality is proved. If $S_{z} \leqslant 0$, we can prove that $S_{y}+S_{z} \geqslant 0$.

In fact,
$$\begin{aligned} & S_{y}+S_{x}=\frac{1}{x z}-\frac{3}{(x+y)(z+y)}+\frac{1}{x y}-\frac{3}{(x+z)(z+y)} \\ = & \frac{-4 y x^{2} z+x^{2} y^{2}+2 y^{2} z^{2}+y^{3} x+y^{3} z+x^{2} z^{2}+x z^{3}+y z^{3}}{x z(x+y)(z+y) y(x+z)} \end{aligned}$$

When $x, y, z \in[1,2]$, the above expression is clearly greater than zero.
Since $(x-z)^{2} \geqslant(x-y)^{2}$, we have
$$\sum_{\text {cyc }} S_{x}(y-z)^{2} \geqslant S_{x}(y-z)^{2}+\left(S_{y}+S_{z}\right)(x-y)^{2} \geqslant 0$$

From equation (10), we know that equation (9) holds, i.e., the original inequality holds, with equality if and only if $x=y=z$ or $x=2, y=z=1$.