Long Huynh Huu's Inequality and Solution


A Simplified Version of Leo Giugiuc's Inequality from the AMM



Let $x_1,...,x_n$ be positive integers, $n \geq 1$. Then


$\displaystyle \sum_{i=1}^n \arctan(x_i) \geq \arctan(x_1...x_n)$

Proof of Lemma

WLOG, $x_1\geq ... \geq x_n$.


$\displaystyle \sum_{i=1}^n \arctan(x_i) \geq \arctan(x_1) \geq \arctan(\sqrt[n]{x_1...x_n}),$

as the maximal value $x_1$ is greater than (or equal to) the geometric mean of all $x_i$.

Case 1: $\mathbf{\displaystyle \prod_{i=1}^n x_i \leq 1}$

In this case $\sqrt[n]{\prod_i x_i} \geq \prod_i x_i$, so the claim follows from (1) and monotonicity of $\arctan$.

Case 2: $\mathbf{\displaystyle \prod_{i=1}^n x_i \gt 1}$

If $x_2 \geq 1$, then

$\displaystyle \arctan(x_1) + \arctan(x_2) \geq \frac{\pi}{4} +\frac{\pi}{4} = \frac{\pi}{2} > \arctan(x_1...x_n).$

Otherwise if $x_1 \gt 1$ and $x_2 \lt 1$, then

$\displaystyle x_2 \gt \prod_{i=2}^{n} x_i \gt \frac{1}{x_1}.$

and hence

$\displaystyle\begin{align} \arctan(x_1) + \arctan(x_2) &\gt \arctan(x_1) + \arctan(1/x_1)\\ &= \frac{\pi}{2} \gt \arctan(x_1...x_n). \end{align}$

In any case, the claim follows.


Using $\displaystyle \arctan(1/x_i)=\frac{\pi}{2} - \arctan(x_i)$ we get

$\displaystyle \begin{align} &\sum_{i=1}^n \arctan(1/x_i) \geq \arctan\left(\frac{1}{x_1...x_n}\right)\\ &\qquad\qquad\iff n\frac{\pi}{2} \geq \sum_{i=1}^n \arctan(x_i) +\arctan\left(\frac{1}{x_1...x_n}\right) \end{align}$


This problem was posed on twitter and solved by Long Huynh Huu.


Inequalities with the Product of Variables as a Constraint

|Contact| |Up| |Front page| |Contents| |Algebra|

Copyright © 1996-2017 Alexander Bogomolny


Search by google: