Why is $\alpha^n\beta^m+\alpha^m\beta^n\le\alpha^s\beta^t+\alpha^t\beta^s$ if $(n,m)\preceq(s,t)$? I've recently come across the following statement:
$$\alpha^n\beta^m+\alpha^m\beta^n\le\alpha^s\beta^t+\alpha^t\beta^s$$
for $\alpha,\beta\ge0$, $n,m,s,t\in\mathbb N_+$, $n+m=s+t$, and $(n,m)\preceq(s,t)$.
If anyone is interested (though it's not particularly relevant for the question itself), I found this stated in a slightly different form in this paper, page 2, second column. They reference a book on majorization in which I wasn't able to find an easy answer to this particular point (both paper and book are paywalled, sorry about that).
I can prove the case $n=m$ (see below), but I'm at a loss for the more general statement. What is an easy way to prove this?
Also, can this sort of statement be generalized to products of more than two numbers? The direct generalization would seem to be
$$\sum_{\sigma\in\mathcal P_N}\prod_{i=1}^N \alpha_{\sigma(i)}^{n_i}\le\sum_{\sigma\in\mathcal P_N}\prod_{i=1}^N \alpha_{\sigma(i)}^{m_i}$$
when $(n_1,..,n_N)\preceq(m_1,...,m_N)$.
If this holds, can the same technique used for the first statement be applied to this more general case?

For $n=m$ we write $s=\mu+\Delta, t=\mu-\Delta$, where $$\mu\equiv\frac{s+t}{2}=\frac{n+m}{2}=n.$$
We thus have
\begin{align}
(\alpha^n\beta^m+\alpha^m\beta^n)-(\alpha^s\beta^t+\alpha^t\beta^s)
&=2(\alpha\beta)^n-(\alpha\beta)^{\mu-\Delta}(\alpha^{2\Delta}+\beta^{2\Delta}) \\
&=(\alpha\beta)^{\mu-\Delta}\left[2\alpha^\Delta\beta^\Delta-\alpha^{2\Delta}-\beta^{2\Delta}\right]\le0,
\end{align}
where the last step follows from
$$2\alpha^\Delta\beta^\Delta-\alpha^{2\Delta}-\beta^{2\Delta}
=-(\alpha^\Delta-\beta^\Delta)^2.$$
The sign of the inequality is consistent with the original statement because
$(n,n)\preceq(n+\Delta,n-\Delta)$ (provided $n-\Delta\ge0$).
 A: The inequality you are interested in is wrong. To see this, take $m=3, n=1, s=t=2$. Then it is easy to see that the following inequality $\alpha^3\beta + \alpha \beta^3 \leq 2 \alpha^2 \beta^2$ does not hold.
In fact, it must have an opposite sign. In this case it is a particular case of a powerful Muirhead's inequality.
Let $a = (a_1,\dots, a_n)$ and $x=(x_1,\dots, x_n)$ be real nonnegative numbers. Define a symmetric polynomial $a\mapsto\Phi(a)$ with variables $x_1,\dots, x_n$ to be an average of all possible terms $x_{\sigma_1}^{a_1}...x_{\sigma_n}^{a_n}$ over all possible permutations $\sigma $ of ${1,\dots, n}$. In other words, it is 
$$\Phi (a) = \frac{1}{n!}\sum_{\sigma}x_{\sigma_1}^{a_1}...x_{\sigma_n}^{a_n}.$$
For example, $\Phi(3,2) = \frac 1 2 (x_1^3x_2^2 + x_1^2x_2^3)$, $\Phi(3,2,0) = \frac 1 6 (x_1^3x_2^2 + x_1^3 x_3^2 + x_1^2x_2^3+x_1^2x_3^3 + x_2^3x_3^2+x_2^2x_3^3)$.
Then Muirhead's inequality says that if $a$ majorizes $b$, then $\Phi(a)\geq \Phi(b)$.
In your case $\Phi(m,n) = \frac 1 2 (\alpha^m \beta^n +\alpha^n \beta^m)$ and similarly for $\Phi(s,t)$. Now since $(n,m)$ majorizes $(s,t)$, then $\Phi (n,m)\geq \Phi(s,t)$.
The generalization works in the same way.
A: I eventually figured out how to prove the simple case, though I'm not sure how to generalize the idea to the more general one.
The main point is to "parametrize" $m,n,s,t$ via their averages, which the assumption $n+m=s+t$ implies must be equal:
\begin{align}
n&=\mu+\Delta,\\
m&=\mu-\Delta,\\
s&=\mu+\delta,\\
t&=\mu-\delta.
\end{align}
With these definitions, $(n,m)\succeq(s,t)$ is equivalent to $\Delta\ge\delta$.
With these definitions we have
\begin{align}
C\equiv (\alpha^n\beta^m+\alpha^m\beta^n)-(\alpha^t\beta^s+\alpha^s\beta^t)
=(\alpha\beta)^{\mu}\left[
(\alpha/\beta)^{\Delta} +
(\alpha/\beta)^{-\Delta} -
(\alpha/\beta)^{\delta} -
(\alpha/\beta)^{-\delta}
\right].
\end{align}
Let me now define $\gamma\equiv\alpha/\beta$ for notational convenience. After some algebraic manipulation we get to
$$C=(\alpha\beta)^\mu\gamma^{-\Delta}(1-\gamma^{\Delta-\delta})(1-\gamma^{\Delta+\delta}).$$
It then follows that for $\Delta\ge\delta$ we have $C\ge0$.
The problem with this is that it relies to a "parametrisation" of $n,m,s,t$, which is likely hard to do in a more general case with sequences with more than two numbers.
