BMO 2019 Shortlist

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 $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots =\frac{a_n}{b_n}$. 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 $$\frac{1}{{\sum }_{j=1}^n{x}_j}+\frac{1}{{\sum }_{j=1}^n{y}_j}\le \frac{1}{{\sum }_{j=1}^n\frac{x_j{y}_j}{x_j+y_j}}\text{or}\sum _{j=1}^n\frac{x_j{y}_j}{x_j+y_j}\le \frac{\left({\sum }_{j=1}^n{x}_j\right)\left({\sum }_{j=1}^n{y}_j\right)}{{\sum }_{j=1}^n{x}_j+{\sum }_{j=1}^n{y}_j}.$$ 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 $$\frac{a_{i1}}{a_{11}}=\frac{a_{i2}}{a_{12}}=\cdots =\frac{a_{in}}{a_{1n}}$$ for all $i=1,2,\dots ,m$. $\square$