TITLE: Is this inequality on sums of powers of two sequences correct?
QUESTION [27 upvotes]: Let $a_1,\dots,a_n$ and $b_1,\dots,b_n$ be two sequences of non negative numbers such that for every positive integer $k$, 
$$ a_1^k+\cdots+a_n^k \leq b_1^k+\cdots+b_n^k,$$
and 
$$a_1+\cdots+a_n =  b_1+\cdots+b_n.$$
Can we conclude
$$\sqrt{a_1}+\cdots+\sqrt{a_n}\geq \sqrt{b_1}+\cdots+\sqrt{b_n}$$

REPLY [22 votes]: As a counter-example with $n=3$
$$a_1=6, a_2=42,a_3=52$$
$$b_1=12, b_2=22, b_3=66$$
Then

$a_1+a_2+a_3 = 100 = b_1+b_2+b_3$
$a_1^p+a_2^p+a_3^p \lt  b_1^p+b_2^p+b_3^p$ for $p \gt 1$
$\sqrt{a_1}+\sqrt{a_2}+\sqrt{a_3} \lt 16.2 \lt \sqrt{b_1}+\sqrt{b_2}+\sqrt{b_3}$ 

though notice that $a_1^p+a_2^p+a_3^p \gt  b_1^p+b_2^p+b_3^p$ for $0.851 \le p \lt 1$, and I suspect all such counterexamples with $n=3$ reverse the inequality in a small interval below $1$ 

Added
Perhaps a more interesting counter-example is 
$$a_1=1, a_2=4,a_3\approx 5.3931524748543$$
$$b_1=2, b_2=2, b_3\approx 6.3931524748543$$
where $ 6.3931524748543$ is an approximation to the solution of $x^x=16 (x-1)^{x-1}$, so $\sum a_i = \sum b_i$ and $\prod a_i^{a_i} = \prod {b_i}^{b_i}$
This has $$a_1^p+a_2^p+a_3^p \le  b_1^p+b_2^p+b_3^p$$ for all non-negative real $p$ (integer or not), and equality only when $p=0$ or $1$