Problem 1818

We will use the following
Lemma. If $a_1, a_2, \dots, a_n, b_1, b_2, \dots, b_n$ are positive real numbers then
$$\frac{1}{\sum_{j=1}^{n} \frac{1}{a_j}} + \frac{1}{\sum_{j=1}^{n} \frac{1}{b_j}} \le \frac{1}{\sum_{j=1}^{n} \frac{1}{a_j + b_j}}.$$
The equality holds when $\frac{a_1}{b_1} = \frac{a_2}{b_2} = \dots = \frac{a_n}{b_n}$.
*Proof.* Set $x_j = \frac{1}{a_j}$ and $y_j = \frac{1}{b_j}$ for each $j = 1, 2, \dots, n$. Then we have to prove that
$$\frac{1}{\sum_{j=1}^{n} x_j} + \frac{1}{\sum_{j=1}^{n} y_j} \le \frac{1}{\sum_{j=1}^{n} \frac{x_j y_j}{x_j + y_j}} \quad \text{or} \quad \sum_{j=1}^{n} \frac{x_j y_j}{x_j + y_j} \le \frac{\left(\sum_{j=1}^{n} x_j\right) \left(\sum_{j=1}^{n} y_j\right)}{\sum_{j=1}^{n} x_j + \sum_{j=1}^{n} y_j}.$$
Subtract $\sum_{j=1}^{n} x_j$, and we have to prove that
$$\sum_{j=1}^{n} \left( x_j - \frac{x_j y_j}{x_j + y_j} \right) \ge \sum_{j=1}^{n} x_j - \frac{\left( \sum_{j=1}^{n} x_j \right) \left( \sum_{j=1}^{n} y_j \right)}{\sum_{j=1}^{n} x_j + \sum_{j=1}^{n} y_j}$$
or
$$\sum_{j=1}^{n} \left( \frac{x_j^2}{x_j + y_j} \right) \ge \frac{\left( \sum_{j=1}^{n} x_j \right)^2}{\sum_{j=1}^{n} x_j + \sum_{j=1}^{n} y_j}.$$
The last one is a consequence of Cauchy-Schwarz inequality and thus the lemma is proved.
We will now prove that repeating the lemma we will get the desired inequality. For example, if $a_1, a_2, \dots, a_n, b_1, b_2, \dots, b_n, c_1, c_2, \dots, c_n$ are positive reals then by repeating lemma two times we get
$$\frac{1}{\sum_{j=1}^{n} \frac{1}{a_j}} + \frac{1}{\sum_{j=1}^{n} \frac{1}{b_j}} + \frac{1}{\sum_{j=1}^{n} \frac{1}{c_j}} \le \frac{1}{\sum_{j=1}^{n} \frac{1}{a_j + b_j}} + \frac{1}{\sum_{j=1}^{n} \frac{1}{c_j}} \le \frac{1}{\sum_{j=1}^{n} \frac{1}{(a_j + b_j) + c_j}} = \frac{1}{\sum_{j=1}^{n} \frac{1}{a_j + b_j + c_j}}.$$

Using similar reasoning we can prove by induction that
$$\sum_{i=1}^{m} \left( \sum_{j=1}^{n} \frac{1}{a_{ij}} \right)^{-1} = \sum_{i=1}^{m} \frac{1}{\sum_{j=1}^{n} \frac{1}{a_{ij}}} \le \frac{1}{\sum_{j=1}^{n} \frac{1}{\sum_{i=1}^{m} a_{ij}}} = \left( \sum_{j=1}^{n} \left( \sum_{i=1}^{m} a_{ij} \right)^{-1} \right)^{-1},$$
which is the desired result.
The equality holds iff
$$\frac{a_{i1}}{a_{11}} = \frac{a_{i2}}{a_{12}} = \dots = \frac{a_{in}}{a_{1n}}$$
for all $i = 1, 2, \dots, m$. $\square$