مسئله ۶۷۶۷ از سری B مجله ریاضی، رومانی // Problem 17 384 from a "Gazeta Matematică seria B" , Romania

 به پست وبلاگ دیگرم هم سر بزنید

I've written about this number before here. Today I'm posting another problem.

                  " 17 384.  Show that if  $a+b+c=abc,\;(a,\;b,\;c \neq 0)$, then 

$a\left (\frac{1}{b}+\frac{1}{c} \right )+b\left (\frac{1}{c}+\frac{1}{a} \right )+c\left ( \frac{1}{a}+\frac{1}{b}\right )+3=ab+bc+ca.$

 Dan SECLEMAN, student, Craiova"

Solution CiP

We will do a calculation in which at some point we make the substitutions

$a+b=abc-c,\;\;\;b+c=abc-a,\;\;\;a+c=abc-b \tag{S}$

 From left to right, we march :   $a(\frac{1}{b}+\frac{1}{c})+b(\frac{1}{c}+\frac{1}{a})+c(\frac{1}{a}+\frac{1}{b})+3=$

$=\underline{\underline{\frac{a}{b}}}+\frac{a}{c}+\frac{b}{c}+\underline{\frac{b}{a}}+\underline{\frac{c}{a}}+\underline{\underline{\frac{c}{b}}}+3=\frac{1}{a}\underline{(b+c)}+\frac{1}{b}\underline{(\underline{a+c})}+\frac{1}{c}(a+b)+3\overset{(S)}{=}$

$\underset{(S)}{=}\frac{1}{a}(abc-a)+\frac{1}{b}(abc-b)+\frac{1}{c}(abc-c)+3=(bc-1)+(ac-1)+ab-1)+3=$

$=ab+bc+ca.$ QED