<s>[INST] For non-negative reals such that $a+b+c\geq x+y+z$, $ab+bc+ca\geq xy+yz+zx$, and $abc\geq xyz$, show $a^k+b^k+c^k\geq x^k+y^k+z^k$ for $0
Let $a$, $b$, $c$, $x$, $y$, $z$ be non-negative real numbers such that
  $$a+b+c \geq x+y+z,$$
$$ab+bc+ca \geq xy+yz+zx,$$
$$ abc \geq xyz$$
  Show that
  $$a^k+b^k+c^k \geq x^k+y^k+z^k, \quad 0 < k < 1$$

This is case $n=3$ of a problem posted by Ji Chen in the Art of Problem Solving forums, 2008.
I have posted a partial result in an answer below.
I have a proof when $r = \frac12,$ for weaker conditons
$$a+b+c = x+y+z,$$
$$\min(x, y, z) \leqslant \min(a, b, c),$$
$$\max(a, b, c) \leqslant \max(x, y, z).$$
Indeed, if $u, v > 0$ it's easy check
$$\sqrt{u} - \sqrt{v} \leqslant \frac{u-v}{2\sqrt{v}}.$$
Assume $x \geqslant y \geqslant z$ and $a \geqslant b \geqslant c$ then $x \geqslant a, \; c \geqslant z.$ Therefore
$$\begin{aligned}\sqrt{x}+\sqrt{y}+\sqrt{z} - \sqrt{a} - \sqrt{b} - \sqrt{c} & \leqslant \frac{x-a}{2\sqrt{a}}+ \frac{y-b}{2\sqrt{b}}+ \frac{z-c}{2\sqrt{c}} \\& \leqslant \frac{x-a}{2\sqrt{b}}+ \frac{y-b}{2\sqrt{b}}+ \frac{z-c}{2\sqrt{c}} \\& =\frac{x+y+z-a-b-c}{2\sqrt{b}}-\frac{(c-z)(\sqrt{b}-\sqrt{c})}{2\sqrt{bc}} \leqslant 0.\end{aligned}$$
 [/INST] Thank @Michael Rozenberg,  I have a proof when $k = \frac12,$ for weaker conditons
$$a+b+c = x+y+z,$$
$$\min(x, y, z) \leqslant \min(a, b, c),$$
$$\max(a, b, c) \leqslant \max(x, y, z).$$
Indeed, if $u, v > 0$ it's easy check
$$\sqrt{u} - \sqrt{v} \leqslant \frac{u-v}{2\sqrt{v}}.$$
Assume $x \geqslant y \geqslant z$ and $a \geqslant b \geqslant c$ then $x \geqslant a, \; c \geqslant z.$ Therefore
$$\begin{aligned}\sqrt{x}+\sqrt{y}+\sqrt{z} - \sqrt{a} - \sqrt{b} - \sqrt{c} & \leqslant \frac{x-a}{2\sqrt{a}}+ \frac{y-b}{2\sqrt{b}}+ \frac{z-c}{2\sqrt{c}} \\& \leqslant \frac{x-a}{2\sqrt{b}}+ \frac{y-b}{2\sqrt{b}}+ \frac{z-c}{2\sqrt{c}} \\& =\frac{x+y+z-a-b-c}{2\sqrt{b}}-\frac{(c-z)(\sqrt{b}-\sqrt{c})}{2\sqrt{bc}} \leqslant 0.\end{aligned}$$
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