Weyl-Hardy-Littlewood-Polya inequality Let $a_1,\cdots,a_n$ and $b_1,\cdots,b_n$ be two decreasing sequences of positive numbers such that $\sum_{k=1}^n a_k\leq \sum_{k=1}^n b_k$ and assume $F(x)$ is a convex function defined on $\mathbb R$ with $\displaystyle\lim_{x\to-\infty}F(x)=0$.
Does it follow (perhaps by Weyl-Hardy-Littlewood or Karamata inequality) that
$$\sum_{k=1}^n F(a_k)\leq \sum_{k=1}^n F(b_k)$$
In particular does the inequality $\displaystyle\sum_{k=1}^n a^2_k\leq
 \sum_{k=1}^n b^2_k$ hold?
 A: (the original question asked about concave $F$ and gave the particular case of whether $\sum \sqrt{a_k} \le \sum \sqrt{b_k}$. This post offered a counter-example to that.)
$0.25 + 0.24 < 0.5 + 0.01$ but $0.5 + 0.4898979... > 0.7071... + 0.1$
Though I note that your particular case does not satisfy the condition that $$\lim_{x\to-\infty} F(x) = 0$$

It took me a some time thinking about the new version before I realized I was overlooking an easy counter: Choose the two sequences so that their sum is equal. Then you can consider either one to be $\{a_k\}$ and the other to be $\{b_k\}$. Unless $F$ is linear, we can expect that in general $\sum F(a_k) \ne \sum F(b_k)$. So choose $\{a_k\}$ to be the sequence that gives the larger value of $\sum F(a_k)$, and your inequality fails.
Further, if $F$ is continuous for at least one of the $a_k$ or $b_k$, then we can make a minute adjustment to that value to obtain a counter-example with strict inequality.
Note also that the only properties of $F$ used are non-linearity and continuity at one point. So the problem here is not $F$, but rather the weakness of the conditions on $\{a_k\}$ and $\{b_k\}$.
