An Inequality with Two Pairs of Triplets


An Inequality with Two Pairs of Triplets

Solution 1

Pure algebra

$\displaystyle \begin{align} &(a^2+b^2+c^2)\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+\frac{2(ab+bc+ca)(x+y+z)}{xyz}\\ &\qquad\qquad=\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+\frac{2ab}{xy}+\frac{2bc}{yz}+\frac{2ca}{zx}\\ &\qquad\qquad\qquad\qquad+\frac{a^2}{y^2}+\frac{b^2}{z^2}+\frac{c^2}{x^2}+\frac{2ab}{yz}+\frac{2bc}{zx}+\frac{2ca}{xy}\\ &\qquad\qquad\qquad\qquad+\frac{a^2}{z^2}+\frac{b^2}{x^2}+\frac{c^2}{y^2}+\frac{2ab}{zx}+\frac{2bc}{xy}+\frac{2ca}{yz}\\ &\qquad\qquad=\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)^2+\left(\frac{a}{y}+\frac{b}{z}+\frac{c}{x}\right)^2+\left(\frac{a}{z}+\frac{b}{x}+\frac{c}{y}\right)^2\\ &\qquad\qquad\ge 0. \end{align}$

Solution 2

$xyz \neq 0$ implies the finiteness of $p=1/x$, $q=1/y$, and $r=1/z$. Thus, the inequality can be written as

$(a^2+b^2+c^2)(p^2+q^2+r^2)+2(ab+bc+ca)(pq+qr+rp)\geq 0.$

The inequality is trivially satisfied if $a^2+b^2+c^2=0$ or $p^2+q^2+r^2=0$. Let us consider the case when neither is zero. The inequality is separately homogeneous in $\{a,b,c\}$ and $\{p,q,r\}$. Thus, WLOG, we can assume $a^2+b^2+c^2=1$ and $p^2+q^2+r^2=1$.

Let us find the extrema of $ab+bc+ca$ under the constraint $a^2+b^2+c^2=1$ using Lagrange multipliers. The three resulting equations obtained in addition to the constraint are

$ b+c-2\lambda a = 0 \\ c+a-2\lambda b = 0 \\ a+b-2\lambda c = 0,$

where $\lambda$ is the Lagrange multiplier. Adding the three equations, we have

$ (a+b+c)(1-\lambda)=0.$

Thus, either $a+b+c=0$ or $\lambda=1$.

$\lambda=1$ results in $a=b=c$. The constraint implies $a=b=c=\pm 1/\sqrt{3}$. Thus, $ab+bc+ca=1$ for this case. If $a+b+c=0$,

$\displaystyle ab+bc+ca=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}=-\frac{1}{2}.$

The exact same analysis applies to $\{p,q,r\}$.

Thus, the LHS can be written as $1+2uv$ where $u\in[-1/2,1]$ and $v\in[-1/2,1]$. This expression will take minimum value when one of $\{u,v\}$ is most negative (takes value $-1/2$) and the other is most positive (takes value $+1$). Thus the minimum value of the LHS is $1+2(-1/2)(1)=0$.


The problem above was kindly posted to the CutTheKnotMath facebook page by Dan Sitaru, with a solution by Ravi Prakash. Originally, the problem was published by Dan at the Romanian Mathematical Magazine. Solution 2 is by Amit Itagi.


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

Copyright © 1996-2018 Alexander Bogomolny