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
$$\sqrt{a_1}+\cdots+\sqrt{a_n}\geq \sqrt{b_1}+\cdots+\sqrt{b_n}$$

3$\begingroup$ Related note: If $(a_1,\ldots,a_n)$ majorizes $(b_1,\ldots,b_n)$, then the conditions are satisfied and the conclusion holds (because $\sum_{1\le i\le n} f(x_i)$ is Schurconvex for convex $f$). Therefore if you are interested some special cases then you can look at majorization. $\endgroup$ – Yuzhou Gu May 31 '18 at 18:54

2$\begingroup$ I think in general this will not hold. See Theorem 1 of link.springer.com/content/pdf/10.1007%2Fs1111700620564.pdf  you need additional conditions. $\endgroup$ – Suvrit May 31 '18 at 19:01

$\begingroup$ I am just an undergrad; is there an intuitive reason why you would expect the inequality to hold given the conditions? $\endgroup$ – Ovi Jun 1 '18 at 2:02
As a counterexample 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 counterexample 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 $x^x=16 (x1)^{x1}$, 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 nonnegative real $p$ (integer or not), and equality only when $p=0$ or $1$

$\begingroup$ Nice! I wonder if there is a $p \in [0.851,1]$ such that no counterexamples exist for any $n$... $\endgroup$ – usul May 31 '18 at 20:39

$\begingroup$ @usul  Perhaps with $n>3$ you could get the derivative of the difference with respect to $p$ to be $0$ when $p=1$, which might need $a_1^{a_1}a_2^{a_2}\cdots a_n^{a_n} = b_1^{b_1}b_2^{b_2}\cdots b_n^{b_n}$ and this might lead to counterexamples where there was no reversal of the inequality  just speculating $\endgroup$ – Henry May 31 '18 at 20:51

3$\begingroup$ Here's a "pure thought" variant of your example. Take $a_1=\epsilon$, $a_2 =12\epsilon$, $a_3=2+\epsilon$; $b_1 =2\epsilon$, $b_2 = 14\epsilon$ and $b_3= 2+2\epsilon$. Think of $\epsilon$ as an arbitrarily small positive quantity. Then it is easy to check that for $p\ge 2$ one has $a_1^p + a_2^p +a_3^p \le b_1^p+ b_2^p + b_3^p$; and the inequality continues to hold for all $0< p <1$ because $(2\epsilon)^p > \epsilon^p$ in that case. $\endgroup$ – Lucia May 31 '18 at 21:29

1$\begingroup$ In fact, $a = 4, 4, 1$ and $b = 5, 2, 2$ also works for any $p < 1$, and doesn't use $0$. $\endgroup$ – tomsmeding May 31 '18 at 21:30

1$\begingroup$ @tomsmeding I do not think your examples work for $p=1.5$  admittedly not an integer, but mine works for all real $p > 1$ $\endgroup$ – Henry May 31 '18 at 21:40
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 231238):

4$\begingroup$ I doubt this will be enough for a counterexample...in the proof, it looks like necessity of condition 1.3 follows from taking $p \to 0$ and of condition 1.5 from taking $p \to \infty$. Neither of these is applicable here, as among all $p < 1$ we only need to take $p = 0.5$. $\endgroup$ – usul May 31 '18 at 19:47

$\begingroup$ Interestingly, all the counterexamples reported so far violate the condition (1.3) of the above theorem. $\endgroup$ – Suvrit Jun 2 '18 at 16:51
Alternatively, for $n=2$ we have:
$$\begin{align} 0=(a_1+a_2)^2(b_1+b_2)^2&=(a_1^2+a_2^2)(b_1^2+b_2^2)+2a_1a_22b_1b_2 \\ &\leq 2a_1a_22b_1b_2 \end{align}.$$ 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 $$\sqrt{a_1}+\sqrt{a_2}\geq \sqrt{b_1}+\sqrt{b_2}.$$
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 rewritten 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 $b_1 b_2 \le a_2$.
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 $0\lt k \lt 1$.

1$\begingroup$ You may have missed the direction of the final inequality in the question? $\endgroup$ – usul May 31 '18 at 19:12