An Inequality Involving Sums of Reciprocals of Pair Sums

Solution 1:

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.

Solution 2:

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.