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 $$\begin{aligned}\sqrt{x}+\sqrt{y}+\sqrt{z} - \sqrt{a} - \sqrt{b} - \sqrt{c} & \leqslant \frac{x-a}{2\sqrt{a}}+ \frac{y-b}{2\sqrt{b}}+ \frac{z-c}{2\sqrt{c}} \\& \leqslant \frac{x-a}{2\sqrt{b}}+ \frac{y-b}{2\sqrt{b}}+ \frac{z-c}{2\sqrt{c}} \\& =\frac{x+y+z-a-b-c}{2\sqrt{b}}-\frac{(c-z)(\sqrt{b}-\sqrt{c})}{2\sqrt{bc}} \leqslant 0.\end{aligned}$$
Let $a$, $b$ and $c$ be different numbers, $a+b+c=u$, $ab+ac+bc=v$ and $abc=w$.
Now, by $u$, $v$ and $w$ we can get any permutations of $a$, $b$ and $c$, but since $a^k+b^k+c^k$ is a symmetric expression, we can think that $a^k+b^k+c^k=f(u,v,w)$ and we need to prove that
$f$ increases as a function of $u$, as a function of $v$ and as a function of $w$.
Id est, it's enough to prove that: $\frac{\partial f}{\partial u}\geq0,$ $\frac{\partial f}{\partial v}\geq0$ and $\frac{\partial f}{\partial w}\geq0.$
Indeed, $$1=\frac{\partial(a+b+c)}{\partial u}=\frac{\partial a}{\partial u}+\frac{\partial b}{\partial u}+\frac{\partial c}{\partial u},$$ $$0=\frac{\partial(ab+ac+bc)}{\partial u}=\frac{\partial a}{\partial u}b+\frac{\partial b}{\partial u}a+\frac{\partial a}{\partial u}c+\frac{\partial c}{\partial u}a+\frac{\partial b}{\partial u}c+\frac{\partial c}{\partial u}b=$$ $$=(b+c)\frac{\partial a}{\partial u}+(a+c)\frac{\partial b}{\partial u}+(a+b)\frac{\partial c}{\partial u}.$$ Also, $$0=\frac{\partial(abc)}{\partial u}=bc\frac{\partial a}{\partial u}+ac\frac{\partial b}{\partial u}+ab\frac{\partial c}{\partial u}.$$ Now, the determinant of this system it's $$\Delta=\left|\left(\begin{array}{cc} 1&1&1\\b+c&a+c&a+b\\bc&ac&ab\end{array}\right)\right|=\sum_{cyc}(ab(a+c)-bc(a+c))=$$ $$=\sum_{cyc}(a^2b-a^2c)=(a-b)(a-c)(b-c);$$ $$\Delta_{\frac{\partial a}{\partial u}}=\left|\left(\begin{array}{cc}1&1&1\\0&a+c&a+b\\0&ac&ab\end{array}\right)\right|=(a+c)ab-ac(a+b)=a^2(b-c),$$ which gives $$\frac{\partial a}{\partial u}=\frac{a^2(b-c)}{(a-b)(a-c)(b-c)}=\frac{a^2}{(a-b)(a-c)}.$$ Similarly, $$\frac{\partial b}{\partial u}=\frac{b^2}{(b-a)(b-c)}$$ and $$\frac{\partial c}{\partial u}=\frac{c^2}{(c-a)(c-b)}.$$ By the same way we can get that $$\left(\frac{\partial a}{\partial v},\frac{\partial b}{\partial v},\frac{\partial c}{\partial v}\right)=\left(-\frac{a}{(a-b)(a-c)},-\frac{b}{(b-a)(b-c)},-\frac{c}{(c-a)(c-b)}\right)$$ and $$\left(\frac{\partial a}{\partial w},\frac{\partial b}{\partial w},\frac{\partial c}{\partial w}\right)=\left(\frac{1}{(a-b)(a-c)},\frac{1}{(b-a)(b-c)},\frac{1}{(c-a)(c-b)}\right).$$ Now, let $a>b>c$.
Thus, by the Lagrange's theorem there is $t\in[b,a],$ for which $$\frac{a^{k+1}-b^{k+1}}{a-b}=(k+1)t^k\geq(k+1)b^k,$$ which gives $$a^{k+1}\geq b^{k+1}+(k+1)b^k(a-b).$$ Also, there is $p\in[c,b],$ for which $$\frac{b^{k+1}-c^{k+1}}{b-c}=(k+1)p^k\leq(k+1)b^k,$$ which says $$c^{k+1}\geq b^{k+1}-(k+1)b^k(b-c).$$ Thus, $$\frac{\partial f}{\partial u}=\sum_{cyc}\frac{\partial f}{\partial a}\frac{\partial a}{\partial u}=\sum_{cyc}\frac{ka^{k-1}\cdot a^2}{(a-b)(a-c)}=k\sum_{cyc}\frac{a^{k+1}}{(a-b)(a-c)}\geq$$ $$\geq k\left(\frac{b^{k+1}+(k+1)b^k(a-b)}{(a-b)(a-c)}+\frac{b^{k+1}}{(b-a)(b-c)}+\frac{b^{k+1}-(k+1)b^k(b-c)}{(c-a)(c-b)}\right)=0.$$ By the same way we can prove that $\frac{\partial f}{\partial v}\geq0$ and $\frac{\partial f}{\partial w}\geq0.$
Now, since our proof is valid for $a\rightarrow b^+$ and for $b\rightarrow c^+$ and since $f$ is a continues function, we have that $f$ increases for any positive $a$, $b$ and $c$, which ends a proof.