# Showing 1/2 3^(2-p)(x+y+z)^(p-1)<=(x^p)/(y+z)+(y^p)/(x+z)+(z^p)/(y+x) By Jensen I got taking p_1=p_2=p_3=1/3, (say x=x_1,y=x_2,z=x_3) (sum p_kx_k)^p<=\sum p_kx^p which means; 3^(1-p)(x+y+z)^p<= x^p+y^p+z^p and if I assume wlog x>=y>=z then 1/(y+z)>=1/(x+z)>=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.

Showing $\frac{1}{2}{3}^{2-p}\left(x+y+z{\right)}^{p-1}\le \frac{{x}^{p}}{y+z}+\frac{{y}^{p}}{x+z}+\frac{{z}^{p}}{y+x}$ with $p>1$, and $x,y,z$ positive
By Jensen I got taking ${p}_{1}={p}_{2}={p}_{3}=\frac{1}{3}$, (say $x={x}_{1},y={x}_{2},z={x}_{3}$)
${\left(\sum {p}_{k}{x}_{k}\right)}^{p}\le \sum {p}_{k}{x}^{p}$ which means;
${3}^{1-p}\left(x+y+z{\right)}^{p}\le {x}^{p}+{y}^{p}+{z}^{p}$
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.
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Cael Cox
Indeed, Jensen works.
Let $x+y+z=3$
Hence, we need to prove that $\sum _{cyc}f\left(x\right)\ge 0$, where $f\left(x\right)=\frac{{x}^{p}}{3-x}-\frac{1}{2}$
${f}^{″}\left(x\right)=\frac{{x}^{p-2}\left(\left(p-2\right)\left(p-1\right){x}^{2}-6p\left(p-2\right)x+9p\left(p-1\right)\right)}{\left(3-x{\right)}^{3}}$
If $p=2$ so ${f}^{″}\left(x\right)>0$
If $p>2$ so since ${f}^{″}\left(0\right)>0$ so since ${f}^{″}\left(0\right)>0$ and $\underset{x\to {3}^{-}}{lim}{f}^{″}\left(x\right)>0$ and $\frac{3p}{p-1}>3$, we see that ${f}^{″}\left(x\right)>0$
If $1 so since ${f}^{″}\left(0\right)>0$ and $\underset{x\to {3}^{-}}{lim}{f}^{″}\left(x\right)>0$, we see that ${f}^{″}\left(x\right)>0$
Thus, your inequality follows from Jensen.
Done
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Ibrahim Rosales
WLOG,
$\begin{array}{}\text{(1)}& x+y+z=1\end{array}$
then by power mean inequality;
$\sum _{cyc}{\left(\frac{1}{3}\left(x+y{\right)}^{-1}\right)}^{-1}\le \left(x+y{\right)}^{\frac{1}{3}}\left(y+z{\right)}^{\frac{1}{3}}\left(x+z{\right)}^{\frac{1}{3}}$
and AM-GM
$\left(x+y{\right)}^{\frac{1}{3}}\left(y+z{\right)}^{\frac{1}{3}}\left(x+z{\right)}^{\frac{1}{3}}\le \frac{1}{3}\left(x+y\right)+\frac{1}{3}\left(y+z\right)+\frac{1}{3}\left(x+z\right)\stackrel{\left(1\right)}{=}\frac{2}{3}$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\ge \frac{9}{2}$
Rearrangement Inequality implies,
$\frac{{x}^{p}}{y+z}+\frac{{y}^{p}}{x+z}+\frac{{z}^{p}}{y+x}\ge \frac{\left({x}^{p}+{y}^{p}+{z}^{p}\right)\cdot \left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)}{3}$
and in general one has ${\left(\sum {p}_{k}{x}_{k}\right)}^{p}\le \sum {p}_{k}{x}^{p}$ thus,
${3}^{p-1}\left({x}^{p}+{y}^{p}+{z}^{p}\right)\ge \left(x+y+z{\right)}^{p}\stackrel{\left(1\right)}{=}\left(x+y+z{\right)}^{p-1}$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{{x}^{p}}{y+z}+\frac{{y}^{p}}{x+z}+\frac{{z}^{p}}{y+x}\ge \frac{\left({x}^{p}+{y}^{p}+{z}^{p}\right)\cdot \frac{9}{2}}{3}\ge \frac{{3}^{1-p}\left(x+y+z{\right)}^{p-1}\cdot \frac{9}{2}}{3}=\frac{1}{2}{3}^{2-p}\left(x+y+z{\right)}^{p-1}$