Showing $\frac{1}{2}{3}^{2-p}(x+y+z{)}^{p-1}\le \frac{{x}^{p}}{y+z}+\frac{{y}^{p}}{x+z}+\frac{{z}^{p}}{y+x}$ with $p>1$, and $x,y,z$ positive

By Jensen I got taking ${p}_{1}={p}_{2}={p}_{3}=\frac{1}{3}$, (say $x={x}_{1},y={x}_{2},z={x}_{3}$)

${(\sum {p}_{k}{x}_{k})}^{p}\le \sum {p}_{k}{x}^{p}$ which means;

${3}^{1-p}(x+y+z{)}^{p}\le {x}^{p}+{y}^{p}+{z}^{p}$

and if I assume wlog $x\ge y\ge z$ then $\frac{1}{y+z}\ge \frac{1}{x+z}\ge \frac{1}{x+y}$

so these $2$ sequences are similarly ordered, and by rearrangement inequality the rest follows ?

I should use power means, but any other solution is also appreciated.

By Jensen I got taking ${p}_{1}={p}_{2}={p}_{3}=\frac{1}{3}$, (say $x={x}_{1},y={x}_{2},z={x}_{3}$)

${(\sum {p}_{k}{x}_{k})}^{p}\le \sum {p}_{k}{x}^{p}$ which means;

${3}^{1-p}(x+y+z{)}^{p}\le {x}^{p}+{y}^{p}+{z}^{p}$

and if I assume wlog $x\ge y\ge z$ then $\frac{1}{y+z}\ge \frac{1}{x+z}\ge \frac{1}{x+y}$

so these $2$ sequences are similarly ordered, and by rearrangement inequality the rest follows ?

I should use power means, but any other solution is also appreciated.