Showing $\frac12 3^{2-p}(x+y+z)^{p-1}\le\frac{x^p}{y+z}+\frac{y^p}{x+z}+\frac{z^p}{y+x}$ 
Showing $\frac12 3^{2-p}(x+y+z)^{p-1}\le\frac{x^p}{y+z}+\frac{y^p}{x+z}+\frac{z^p}{y+x}$ with $p>1$, and $x,y,z$ positive 

By Jensen I got taking $p_1=p_2=p_3=\frac13$, (say $x=x_1,y=x_2,z=x_3$)
$\left(\sum p_kx_k\right)^p\le\sum p_kx^p$ 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.
 A: Indeed, Jensen works.
Let $x+y+z=3$.
Hence, we need to prove that $\sum\limits_{cyc}f(x)\geq0$, where $f(x)=\frac{x^p}{3-x}-\frac{1}{2}$.
$f''(x)=\frac{x^{p-2}\left((p-2)(p-1)x^2-6p(p-2)x+9p(p-1)\right)}{(3-x)^3}$.
If $p=2$ so $f''(x)>0$.
If $p>2$ so since $f''(0)>0$ and $\lim\limits_{x\rightarrow3^-}f''(x)>0$ and $\frac{3p}{p-1}>3$, we see that $f''(x)>0$.
If $1<p<2$ so since $f''(0)>0$ and $\lim\limits_{x\rightarrow3^-}f''(x)>0$, we see that $f''(x)>0$.
Thus, your inequality follows from Jensen.
Done!
A: WLOG, $x+y+z=1\tag1$ then by power mean inequality;
$\sum\limits_{cyc}\left(\frac13(x+y)^{-1}\right)^{-1}\le(x+y)^{\frac 13}(y+z)^{\frac13}(x+z)^{\frac1 3}$
and AM-GM
$(x+y)^{\frac 13}(y+z)^{\frac13}(x+z)^{\frac13}\le\frac13(x+y)+\frac13(y+z)+\frac13(x+z)\overset{(1)}=\frac23$
$\displaystyle\implies \frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\ge\frac92$
Rearrangement Inequality implies,
$\displaystyle\frac{x^p}{y+z}+\frac{y^p}{x+z}+\frac{z^p}{y+x}\ge\frac{(x^p+y^p+z^p)\cdot \left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)}{3}$
and in general one has $\left(\sum p_kx_k\right)^p\le\sum p_kx^p$ thus,
$3^{p-1}(x^p+y^p+z^p)\ge(x+y+z)^{p}\overset{(1)}=(x+y+z)^{p-1}$
$\displaystyle\implies \frac{x^p}{y+z}+\frac{y^p}{x+z}+\frac{z^p}{y+x}\ge\frac{(x^p+y^p+z^p)\cdot\frac92}{3}\ge\frac{3^{1-p}(x+y+z)^{p-1}\cdot\frac92}{3}=\frac12 3^{2-p}(x+y+z)^{p-1}$
