2
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ Mike Ernest proved it (see accepted answer) using integration. Is there a pre-calculus proof? $\endgroup$
    – aaa acb
    Nov 23, 2021 at 18:34

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Nice. From the proof it follows that the inequality holds iff the bracket is 0 for all $t\in[0,1]$ iff $\{x_1, \ldots, x_n\} = \{y_1, \ldots, y_m\}$ (multi-set). $\endgroup$
    – aaa acb
    Nov 23, 2021 at 5:16
  • $\begingroup$ It is very nice. (+1) $\endgroup$
    – River Li
    Nov 24, 2021 at 1:44
2
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .