Can You Solve This Intriguing Real Numbers Equation?

[anemone]

Here is this week's POTW:

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Let $a,\,b,\,c$ be real numbers all different from $-1$ and $1$ such that $a+b+c=abc$.

Prove that $\dfrac{a}{1-a^2}+\dfrac{b}{1-b^2}+\dfrac{c}{1-c^2}=\dfrac{4abc}{(1-a^2)(1-b^2)(1-c^2)}$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution:):

1. kaliprasad
2. greg1313
3. lfdahl

Solution from kaliprasad:

Spoiler

Putting $a = \tan\, A, b = \tan\, B, c = \tan\, C$

We see that we have $a + b + c = abc$.

=> $\tan\, A + \tan\, B + \tan\, B = \tan\, A \tan\, B \tan\, C$
∴ $\tan ( A+B+C ) = \dfrac{\tan\, A + \tan\, B + \tan\, C -\tan\, A \tan B \tan C}{1 - \tan\, A \tan\, B - \tan\, B \tan\, C - \tan C \tan A} = 0$
∴ $A+B+C = n\pi$ for integral $n$.

$2A + 2B +2C = 2n\pi$

∴ $tan ( 2A+ 2B+ 2C ) = \tan 2n\pi = 0$

∴ $\tan 2A + \tan 2B + \tan 2C =\tan 2A \tan 2B \tan 2C$
∴ $\dfrac{2a}{1-a^2} + \dfrac{2b}{1-b^2}+ \dfrac{2c}{1-c²} = \dfrac{2a}{1-a^2} * \dfrac{2b}{1-b^2}* \dfrac{2c}{1-c²} = \dfrac{8abc}{(1-a^2)(1-b^2)(1-c^2)}$
∴ $\dfrac{a}{1-a^2} + \dfrac{b}{1-b^2}+ \dfrac{c}{1-c^2} = \dfrac{4abc}{(1-a^2)(1-b^2)(1-c^2)}$

Alternate solution from lfdahl:

Spoiler

I start with the LHS and choose a common denominator: $(1-a^2)(1-b^2)(1-c^2)$:

\[\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}
= \frac{a(1-b^2)(1-c^2)+b(1-a^2)(1-c^2)+c(1-a^2)(1-b^2)}{(1-a^2)(1-b^2)(1-c^2)}\]

If I can show, that the nominator equals: $4abc$, I´m done. Along the way, I make use of the relation:
$a+b+c = abc$ (cf. the underbraces).

\[a(1-b^2)(1-c^2)+b(1-a^2)(1-c^2)+c(1-a^2)(1-b^2) \\\\ =a(1-(b^2+c^2)+b^2c^2)+b(1-(a^2+c^2)+a^2c^2)+c(1-(a^2+b^2)+a^2b^2) \\\\=a - a(b^2+c^2) + abc(bc)+ b - b(a^2+c^2)+abc(ac)+c-c(a^2+b^2)+abc(ab) \\\\=\underbrace{a+b+c}_{=abc}+abc(bc)+abc(ac)+abc(ab)-((ab)b+(ac)c+(ab)a+(bc)c+(ac)a+(bc)b) \\\\=abc+abc(bc)-b(bc)-c(bc)+abc(ac)-a(ac)-c(ac)+abc(ab)-a(ab)-b(ab) \\\\= abc+bc\underbrace{(abc-b-c)}_{=a}+ac\underbrace{(abc-a-c)}_{=b}+ab\underbrace{(abc-a-b)}_{=c} \\\\ = 4abc\]

Thus equality holds in the expression:

\[\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2} = \frac{4abc}{(1-a^2)(1-b^2)(1-c^2)}\]

whenever the three numbers $a,b$ and $c$ obey the relation: $a+b+c = abc$ and $a,b,c \notin \left \{ \pm 1 \right \}$.