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

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    $\begingroup$ 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 Schur-convex 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
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    $\begingroup$ I think in general this will not hold. See Theorem 1 of link.springer.com/content/pdf/10.1007%2Fs11117-006-2056-4.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 counter-example with $n=3$

$$a_1=6, a_2=42,a_3=52$$

$$b_1=12, b_2=22, b_3=66$$


  • $a_1+a_2+a_3 = 100 = b_1+b_2+b_3$
  • $a_1^p+a_2^p+a_3^p \lt b_1^p+b_2^p+b_3^p$ 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 $1$


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 $1$

  • $\begingroup$ 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 counter-examples where there was no reversal of the inequality - just speculating $\endgroup$
    – Henry
    May 31 '18 at 20:51
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    $\begingroup$ Here's a "pure thought" variant of your example. Take $a_1=\epsilon$, $a_2 =1-2\epsilon$, $a_3=2+\epsilon$; $b_1 =2\epsilon$, $b_2 = 1-4\epsilon$ 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
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    $\begingroup$ 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
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    $\begingroup$ @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 231-238):

enter image description here

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    $\begingroup$ 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:

$$\begin{align} 0=(a_1+a_2)^2-(b_1+b_2)^2&=(a_1^2+a_2^2)-(b_1^2+b_2^2)+2a_1a_2-2b_1b_2 \\ &\leq 2a_1a_2-2b_1b_2 \end{align}.$$ 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 $$\sqrt{a_1}+\sqrt{a_2}\geq \sqrt{b_1}+\sqrt{b_2}.$$


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 $b_1 b_2 \le a_2$.



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

enter image description here


What about $a_1=a_2=\frac12,b_1=0,b_2=1?$


  • $a_1^k+a_2^k \lt b_1^k+b_2^k$ when $k \gt 1$.
  • $a_1+a_2=b_1+b_2$


  • $a_1^k+a_2^k \gt b_1^k+b_2^k$ 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

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