An Inequality Involving Sums of Reciprocals of Pair Sums Let $x_1, \ldots, x_n, y_1, \ldots, y_m$ be positive real numbers. Prove or disprove the following inequality.
\begin{equation}
\sum_{i=1}^n\sum_{j=1}^n \frac{1}{x_i+x_j} + \sum_{i=1}^m\sum_{j=1}^m \frac{1}{y_i+y_j} \geq 2\sum_{i=1}^n \sum_{j=1}^m \frac{1}{x_i + y_j}
\end{equation}
Remark. Computer experiments seem to suggest the positive. If this is true, (seems to me) should have been discovered in the past. By (or named after) whom and where to look for references?
 A: It is true. It can be derived from this inequality:
$$
\int_0^1\left(t^{x_1-1/2}+t^{x_2-1/2}+\dots+t^{x_n-1/2}-t^{y_1-1/2}-t^{y_2-1/2}-\cdots-t^{y_m-1/2}\right)^2\,dt\ge 0
$$
Simply expand out the square, resulting in terms like $+t^{x_i+x_j-1},+t^{y_i+y_j-1}$ and $-t^{x_i+y_j-1}$, whose integrals from $0$ to $1$ are $\frac1{x_i+x_j},\frac1{y_i+y_j},$ and $\frac{-1}{x_i+y_j}$, respectively. Move the mixed terms to the other side of the inequality, and the result is proven.
A: Remarks: There are some widely used tricks for this kind of problems. Here is one of them.
Alternative proof:
Using the identity ($q > 0$)
$$\frac{1}{q} = \int_0^\infty \mathrm{e}^{-qt} \,\mathrm{d} t,$$
we have
$$\sum_{i=1}^n\sum_{j=1}^n \frac{1}{x_i + x_j}
= \int_0^\infty \sum_{i=1}^n\sum_{j=1}^n
\mathrm{e}^{-(x_i + x_j)t} \,\mathrm{d} t
= \int_0^\infty (\mathrm{e}^{- tx_1}
+ \cdots + \mathrm{e}^{- tx_n})^2 \,\mathrm{d} t.$$
Similarly, we have
$$\sum_{i=1}^m \sum_{j=1}^m
\frac{1}{y_i + y_j}
= \int_0^\infty (\mathrm{e}^{- ty_1}
+ \cdots + \mathrm{e}^{- ty_m})^2 \,\mathrm{d} t,$$
and
$$\sum_{i=1}^n \sum_{j=1}^m \frac{1}{x_i + y_j}
= \int_0^\infty (\mathrm{e}^{- tx_1}
+ \cdots + \mathrm{e}^{- tx_n})
(\mathrm{e}^{- ty_1}
+ \cdots + \mathrm{e}^{- ty_m}) \,\mathrm{d} t.$$
Thus, we have
$$\mathrm{LHS} - \mathrm{RHS}
= \int_0^\infty (\mathrm{e}^{- tx_1}
+ \cdots + \mathrm{e}^{- tx_n}
- \mathrm{e}^{- ty_1}
- \cdots - \mathrm{e}^{- ty_m})^2 \,\mathrm{d} t \ge 0.$$
We are done.
