Interesting Inequality With Exponents And Base > 1

Question

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.

Answer 1

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

Answer 2

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.