Is this inequality on sums of powers of two sequences correct? 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
$$\sqrt{a_1}+\cdots+\sqrt{a_n}\geq \sqrt{b_1}+\cdots+\sqrt{b_n}$$
 A: I think in general, the claim may not hold without additional conditions. In particular, the following theorem may help obtain a counterexample (see: "An Inequality from Moment Theory" by G. Bennett; Positivity, 11 (2007), pp 231-238): 

A: Alternatively, for $n=2$ we have: 
$$\begin{align}
0=(a_1+a_2)^2-(b_1+b_2)^2&=(a_1^2+a_2^2)-(b_1^2+b_2^2)+2a_1a_2-2b_1b_2 \\
&\leq 2a_1a_2-2b_1b_2 \end{align}.$$
So, $a_1a_2\geq b_1b_2$ and $2\sqrt{a_1a_2}\geq 2\sqrt{b_1b_2}$. Again, since $a_1+a_2=b_1+b_2$ we gather that $a_1+a_2+2\sqrt{a_1a_2}\geq b_1+b_2+2\sqrt{b_1b_2}$. Therefore, $(\sqrt{a_1}+\sqrt{a_2})^2\geq(\sqrt{b_1}+\sqrt{b_2})^2$ and thus 
$$\sqrt{a_1}+\sqrt{a_2}\geq \sqrt{b_1}+\sqrt{b_2}.$$
A: For $n = 2$, the answer is yes. Here is a proof.
Because of the first condition, we have $\max_i(a_i) \le \max_i(b_i)$. Hence, if $a_1 = 0$, it follows from the identity $a_1 + a_2 = b_1 + b_2$ that $b_1 = 0$ or $b_2 = 0$, so that the result obviously holds.
If $a_1 \neq 0$, we can assume WLOG, that $a_1 = 1$. We can also assume that $b_1 = \max(b_1, b_2)$, so that $b_1 \ge 1$. Let us set $p(x) = x(1 + a_2 - x)$.
If $a_2 \le 1$, then $b_1 \ge 1 \ge \frac{1 + a_2}{2}$ and we have $b_1 b_2 = p(b_1) \le p(1)  = a_2$. Hence $b_1 b_2 \le a_2$, which can be re-written as $1 + \sqrt{a_2} \ge \sqrt{b_1} + \sqrt{b_2}$.
If $a_2 \ge 1$, then the identity $1 + a_2 = b_1 + b_2$ implies that $b_2 \le 1$. Therefore $b_2(1 - b_2) \le a_2(1 - b_2)$, which is equivalent to $p(b_2) \le a_2$, that is $b_1 b_2 \le a_2$.
A: 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 $1$
A: CORRECTED ANSWER
Funny, it took me several tries to get the inequality to reverse. I was looking at examples where $\sum a_i=\sum b_i$ as requested and also $\sum a_i^2=\sum b_i^2.$
Two well known examples are 


*

*$0,4,5$ vs $1,2,6$ and 

*$0,3,5,6$ vs $1,2,4,7$


In both cases $\sum a_i^t \lt \sum b_i^t$ outside of $[1,2].$
Here are the graphs of $\sum a_i^t - \sum b_i^t$ in the two cases


OLD BAD ANSWER
What about $a_1=a_2=\frac12,b_1=0,b_2=1?$
Then  
  
  
*
  
*$a_1^k+a_2^k \lt b_1^k+b_2^k$ when $k \gt 1$.
  
*$a_1+a_2=b_1+b_2$
  
  
  but
  
  
*
  
*$a_1^k+a_2^k \gt b_1^k+b_2^k$ when $0\lt k \lt 1$.
  

