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  

$$\sqrt{a\_1}+\cdots+\sqrt{a\_n}\geq \sqrt{b\_1}+\cdots+\sqrt{b\_n}$$



Is this inequality on sums of powers of two sequences correct?
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 $1$ 




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*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 $1$


