A rearrangement inequality for exponentiation function

Update: A year ago, but the first answer is not clear with me. I bounty this question again.

My question: I am looking for a proof or counterexample to the following inequality:

If $$n \in \mathbb{N}$$, and $$a_1 \ge a_2 \ge \cdots \ge a_n \ge 0$$ and $$\alpha_1 \ge \alpha_2 \ge \cdots \ge \alpha_n \ge 0$$ then

$${\left(\sum_{i=1}^{n}{a_i^{\alpha_i}} \right)}^n \ge \prod_{i=1}^{n}{\left(\sum_{j=1}^{n}{a_i^{\alpha_j}} \right)}$$

If $$n \in \mathbb{N}$$ and $$a_1 \ge a_2 \ge \cdots \ge a_n \ge 0$$ and $$0 \le \alpha_1 \le \alpha_2 \le \cdots \le \alpha_n$$ then

$${\left(\sum_{i=1}^{n}{a_i^{\alpha_i}} \right)}^n \le \prod_{i=1}^{n}{\left(\sum_{j=1}^{n}{a_i^{\alpha_j}} \right)}$$

Moreover equality holds in either case if and only if $$a_1=a_2=a_3=...=a_n$$.

[The original question claimed a proof when $$n=2$$ or $$n=3$$.]

• I edited your question to make it a bit clearer. At the end you restated the inequalities for $n=2$ and $n=3$, but I didn't see a proof. So I deleted the $n=3$ case and added the details for $n=2$. – Mark Wildon Jun 19 at 9:45
• I think, the difference for $n=2$ is not $a_1^{\alpha_1}a_2^{\alpha_2} - a_1^{\alpha_2}a_2^{\alpha_1}$. – Fedor Petrov Jun 19 at 10:48

The first inequality is not true for $$n=2$$, for example as =

0.0912    0.2256


alphas =

0.8281    1.7010


LHS =

0.0471


RHS =

0.0574


found by the following Matlab/Octave program:

n=2;

for iter=1:1e4 as=cumsum(rand(1,n)); alphas=cumsum(rand(1,n));

LHS=sum(as.^alphas)^n; RHS=prod(sum(as.^(alphas')));

if LHS

as alphas LHS RHS

end

end

Let $a_1 \geq a_2 \geq \ldots a_n \geq 1$ and $r_1 \geq r_2 \geq \ldots r_n \geq 0$.

Then over all permutations $\pi$ of the $r_j$, the sum $\sum_{i=1}^{n}a_i^{r_{\pi_i}}$ is maximized when $r_i$ are in the same order as the $a_i$. The proof is similar to any standard proof of the original re-arrangment inequality.

For contradiction, suppose that some other permutation is larger. Then there exist terms $a^b$ and $c^d$ in the sum such that $a \geq c$ and $b \leq d$. But we have $a^d+c^b \geq a^b+c^d$, i.e. swapping these two exponents does not decrease the value of the sum.

To prove the inequality, re-write as $(a/c)^b \geq \dfrac{c^{d-b}-1}{a^{d-b}-1}$ and notice that the LHS is at least 1 and the RHS is at most 1.

Note: You need a weaker result and it may be sufficient to assume all the $a_i$s non-negative.

• I don't think so. Do You prove that $a^d+c^b \ge a^c+a^b$ ? – Đào Thanh Oai Jul 20 '18 at 11:09
• The answer above would be correct if each factor on the right-hand side was of the form $\sum_{j=1}^n a_{\sigma(j)}^{\alpha_j}$ for some permutation $\sigma$. But in fact $a_i$ is constant within each factor. – Mark Wildon Jun 19 at 9:20

For the first inequality, it follows (when $$a_n\geqslant 1$$) from Aravind's claim that $$\sum_i a_i^{\alpha_i}$$ is not less than arithmetic mean of $$b_1,\ldots,b_n$$, where $$b_i=\sum_j a_i^{\alpha_j}$$. Thus not less than geometric mean too.

The second inequality looks false by trivial reasons: if $$a_n=0$$ and all $$\alpha_i$$'s are positive, RHS equals 0 while LHS not necessary.

• is the first inequality true if we let $a_n \ge 1$ ? – Đào Thanh Oai Jun 19 at 11:19
• Yes, I had to mention this. – Fedor Petrov Jun 19 at 11:42