I had trouble proving the following inequality:

$\beta > 1$

$(\alpha_{1}\beta^{2\alpha_{1}} + \ldots + \alpha_{n}\beta^{2\alpha_{n}})(\beta^{\alpha_{1}} + \ldots + \beta^{\alpha_{n}}) \geq (\alpha_{1}\beta^{\alpha_1} + \ldots + \alpha_{n}\beta^{\alpha_n})(\beta^{2\alpha_{1}} +\ldots + \beta^{2\alpha_n}) $

I tried using rearrangement inequality but that didn't get me anywhere. I'm not entirely sure how to proceed here.


$$\sum_{k=1}^n\alpha_k\beta^{2\alpha_k}\sum_{k=1}^n\beta^{\alpha_k}-\sum_{k=1}^n\alpha_k\beta^{\alpha_k}\sum_{k=1}^n\beta^{2\alpha_k}=$$ $$=\sum_{ 1\leq k<i\leq n}\left(\alpha_k\beta^{2\alpha_k+\alpha_i}+\alpha_i\beta^{2\alpha_i+\alpha_k}-\alpha_k\beta^{\alpha_k+2\alpha_i}-\alpha_i\beta^{\alpha_i+2\alpha_k}\right)=$$ $$=\sum_{1\leq k<i\leq n}(\alpha_k-\alpha_i)\left(\beta^{2\alpha_k+\alpha_i}-\beta^{\alpha_k+2\alpha_i}\right)=$$ $$=\sum_{1\leq k<i\leq n}\beta^{\alpha_k+2\alpha_i}(\alpha_k-\alpha_i)\left(\beta^{\alpha_k-\alpha_i}-1\right)\geq0.$$

  • $\begingroup$ (+1) Spoilsport! :) $\endgroup$ – Calum Gilhooley Mar 24 at 2:08
  • $\begingroup$ Essentially the same argument proves, I think, that if a set of numbers $x_i$ and two sets of weights $w_i, w'_i$ are such that: $$\frac{w_i}{w_j} \geqslant \frac{w'_i}{w'_j}$$ whenever $x_i > x_j$, then: $$\frac{\sum_iw_ix_i}{\sum_iw_i} \geqslant \frac{\sum_iw'_ix_i}{\sum_iw'_i}.$$ Cf. Mateu Sbert and Jordi Poch, "A necessary and sufficient condition for the inequality of generalized weighted means", Journal of Inequalities and Applications (2016), p.292ff. - although their argument doesn't seem very clear to me, and it seems much simpler and clearer to do what you did in this answer. $\endgroup$ – Calum Gilhooley Mar 24 at 3:08

This argument may well be unnecessarily complicated, but perhaps it will give someone an idea for a shorter proof. It applies a general recipe I extracted from the proof of Lemma 1 in Marjanović and Kadelburg, "A proof of Chebyshev's inequality", The Teaching of Mathematics X, no. 2 (2007), p.107f.

Let $n$ be a positive integer (but the case $n = 1$ is trivial, so we may choose to assume $n \geqslant 2$). Let $u_1, u_2, \ldots, u_n$ and $v_1, v_2, \ldots, v_n$ and $y_1, y_2, \ldots, y_n$ be any real numbers, subject only to the constraint that $y_1 \geqslant y_2 \geqslant \cdots \geqslant y_n$. For $m = 0, 1, 2, \ldots, n$, let $U_m = \sum_{k=1}^m u_k$ and $V_m = \sum_{k=1}^m v_k$. Then $U_0 = V_0 = 0$, and: \begin{align*} & \quad \biggl(\sum_{k=1}^n u_ky_k\biggr)\biggl(\sum_{k=1}^n v_k\biggr) - \biggl(\sum_{k=1}^n u_k\biggr)\biggl(\sum_{k=1}^n v_ky_k\biggr) \\ & = \sum_{k=1}^n (u_kV_n - U_nv_k)y_k = \sum_{m=1}^n (u_mV_n - U_nv_m)y_m \\ & = \sum_{m=1}^n ((U_m - U_{m-1})V_n - U_n(V_m - V_{m-1}))y_m \\ & = \sum_{m=1}^n (U_mV_n - U_nV_m)y_m - \sum_{m=1}^n (U_{m-1}V_n - U_nV_{m-1})y_m \\ & = \sum_{m=1}^{n-1} (U_mV_n - U_nV_m)(y_m - y_{m+1}). \end{align*} If all of the factors $U_mV_n - U_nV_m$ are non-negative, then so is the whole sum. Conversely, if the sum is non-negative for all decreasing sequences $y_1, y_2, \ldots, y_n$, then by taking $y_1 = y_2 = \cdots = y_m = 1$ and $y_{m+1} = y_{m+2} = \cdots = y_n = 0$, we find that $U_mV_n - U_nV_m \geqslant 0$ for $m = 1, 2, \ldots, n - 1$ (trivially also for $m = n$). Thus: \begin{multline*} \biggl(\sum_{k=1}^n u_ky_k\biggr)\biggl(\sum_{k=1}^n v_k\biggr) \geqslant \biggl(\sum_{k=1}^n u_k\biggr)\biggl(\sum_{k=1}^n v_ky_k\biggr) \text{ for all } y_1 \geqslant y_2 \geqslant \cdots \geqslant y_n \\ \iff \biggl(\sum_{k=1}^m u_k\biggr)\biggl(\sum_{k=1}^n v_k\biggr) \geqslant \biggl(\sum_{k=1}^n u_k\biggr)\biggl(\sum_{k=1}^m v_k\biggr) \quad \text{for } m = 1, 2, \ldots, n - 1. \end{multline*} The right hand side can be rewritten as: \begin{equation*} \biggl(\sum_{k=1}^m u_k\biggr)\biggl(\sum_{k=m+1}^n v_k\biggr) \geqslant \biggl(\sum_{k=m+1}^n u_k\biggr)\biggl(\sum_{k=1}^m v_k\biggr) \quad \text{for } m = 1, 2, \ldots, n - 1. \end{equation*} Putting it yet another way: \begin{equation*} \sum_{\substack{1 \leqslant k \leqslant m \\ m < l \leqslant n}} (u_kv_l - v_ku_l) \geqslant 0 \quad \text{for } m = 1, 2, \ldots, n - 1. \end{equation*}

For the present application, first arrange the $\alpha_k$ in non-increasing order (which we can do without loss of generality). Because $\beta > 1$, the terms $\beta^{\alpha_k}$ will also be in non-increasing order.

Make the following assignments, for $k = 1, 2, \ldots, n$: \begin{align*} u_k & = \alpha_k\beta^{\alpha_k}, \\ v_k & = \beta^{\alpha_k}, \\ y_k & = \beta^{\alpha_k}. \end{align*} Then, for all $k$ and $l$ such that $1 \leqslant k \leqslant m < l \leqslant n$: $$ u_kv_l - v_ku_l = (\alpha_k - \alpha_l)\beta^{\alpha_k + \alpha_l} \geqslant 0, $$ so the necessary and sufficient condition on the $u_k$ and $v_k$ is satisfied, and we can apply the above theorem to the $y_k$, obtaining: $$ \biggl(\sum_{k=1}^n u_ky_k\biggr)\biggl(\sum_{k=1}^n v_k\biggr) \geqslant \biggl(\sum_{k=1}^n u_k\biggr)\biggl(\sum_{k=1}^n v_ky_k\biggr), $$ or, in terms of the present problem: $$ \biggl(\sum_{k=1}^n \alpha_k\beta^{2\alpha_k}\biggr) \biggl(\sum_{k=1}^n \beta^{\alpha_k}\biggr) \geqslant \biggl(\sum_{k=1}^n \alpha_k\beta^{\alpha_k}\biggr) \biggl(\sum_{k=1}^n \beta^{2\alpha_k}\biggr). $$ The ease with which that worked out suggests that the heavy machinery didn't need to be applied at all - there's almost bound to be a simpler proof.

  • $\begingroup$ Thank you for the upvote, and the acceptance, but I think Michael Rozenberg's answer is unquestionably the one that should be accepted (unless there's a mistake in it, but I don't think there is). You can easily "unaccept" my answer and accept his (or wait for at least a day or so before accepting any answer - which is always a good policy, I think). $\endgroup$ – Calum Gilhooley Mar 24 at 2:24
  • $\begingroup$ Alright, no problem. I saw that your answer was first, and since both proofs were correct, I thought it would be best to accept yours. I'll change it. $\endgroup$ – Rehaan Ahmad Mar 24 at 2:32
  • 1
    $\begingroup$ See What should I do when someone answers my question?: "Accepting an answer is not mandatory; do not feel compelled to accept the first answer you receive. Wait until you receive an answer that answers your question well."' Personally, I don't think I've ever known a case where there was so little to choose between two answers that the choice came down to which was posted first! (Partly this is because posting an answer similar to one that has gone before is frowned upon.) $\endgroup$ – Calum Gilhooley Mar 24 at 3:19

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