Interesting Inequality With Exponents And Base > 1 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. 
 A: $$\sum_{k=1}^n\alpha_k\beta^{2\alpha_k}\sum_{k=1}^n\beta^{\alpha_k}-\sum_{k=1}^n\alpha_k\beta^{\alpha_k}\sum_{k=1}^n\beta^{2\alpha_k}=$$
$$=\sum_{ 1\leq k<i\leq n}\left(\alpha_k\beta^{2\alpha_k+\alpha_i}+\alpha_i\beta^{2\alpha_i+\alpha_k}-\alpha_k\beta^{\alpha_k+2\alpha_i}-\alpha_i\beta^{\alpha_i+2\alpha_k}\right)=$$
$$=\sum_{1\leq k<i\leq n}(\alpha_k-\alpha_i)\left(\beta^{2\alpha_k+\alpha_i}-\beta^{\alpha_k+2\alpha_i}\right)=$$
$$=\sum_{1\leq k<i\leq n}\beta^{\alpha_k+2\alpha_i}(\alpha_k-\alpha_i)\left(\beta^{\alpha_k-\alpha_i}-1\right)\geq0.$$
A: 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 Marjanovi&cacute;
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.
