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If $a^x=b\,c$, $b^y=c\,a$, and $c^z=a\,b$, then how can I prove that $\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=2$? Thank you. I tried a few approaches but failed and as such there is no work to show.

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This is not true. If $a=b=c=x=y=z=1$, then your precondition is met. But then the sum is $\frac32$. –  alex.jordan Nov 3 '11 at 17:45
    
Yes!But what do you think happens when x,y and z are not equal to 1? –  Eisen Nov 3 '11 at 17:49
    
If $a=b=c=1$, your precondition is met automatically, no matter what $x$, $y$, and $z$ are. So the sum will not be constant. For large $x$, $y$, and $z$, the sum will be close to $3$. –  alex.jordan Nov 3 '11 at 17:52
    
Thank you very much. –  Eisen Nov 3 '11 at 17:55

2 Answers 2

up vote 1 down vote accepted

As has been pointed out by @alex.jordan, the result fails if $a=b=c=1$. We will prove the desired result, for positive $a$, $b$, and $c$. Naturally, there will be a little gap. But let's just barge ahead.

Let $a=e^u$, $b=e^v$, and $c=e^w$. (It is more comforting to work with addition.)
Then from $a^x=bc$ we get $e^{ux}=e^ve^w$, and therefore $ux=v+w$. In a similar way, we get the full system $$ux=v+w;\qquad vy=u+w;\qquad wz=u+v.$$ Add up: $$ux+vy+wz=2(u+v+w).$$ Divide by $u+v+w$: $$\frac{ux}{u+v+w}+ \frac{vy}{u+v+w}+\frac{wz}{u+v+w}=2. \qquad(\ast)$$ Now from $ux=v+w$ we obtain $ux+u=u+v+w$, so $\frac{u}{u+v+w}=\frac{1}{x+1}$. Thus the first term in $(\ast)$ is $\frac{x}{x+1}$. We get similar expressions for the other two terms, and the desired result follows.

Now what about the counterexample! We divided by $u+v+w$, which is not possible if $u+v+w=0$, that is, if $abc=1$. So the amended question could, for example, specify that $a$, $b$, and $c$ are positive and $abc\ne 1$.

We look in more detail at the case $abc=1$. From $a^x=bc$, by multiplying by $a$, we get $a^{x+1}=1$. Similarly, $b^{y+1}=1$ and $c^{z+1}=1$. If any of $a$, $b$, $c$ is different from $1$, say $a$, then $x+1=0$, and the expression in the problem does not make sense. If all of $a$, $b$, $c$ are $1$, then $x$, $y$, and $z$ are unconstrained, so the equality need not hold.

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Thank you very much. –  Eisen Nov 3 '11 at 18:05

Though Andre's answer is completely fine (and better than what I am going to write), assuming you know about Logarithm, and its basic formulas, below is an alternate answer (I am writing this answer as sometime this question appears logarithm chapters of textbooks, of course with additional mention of $a,b,c>0$ and $abc\ne1$):

From the first given equation,

$a^x=bc\Rightarrow \frac{x}{1}=\log_a(bc)=\log_a(abc)-1=\frac{1}{\log_{abc}{a}}-1=\frac{1-\log_{abc}{a}}{\log_{abc}{a}}$

$\Rightarrow \frac{x}{1+x}=\frac{1-\log_{abc}{a}}{\log_{abc}{a}+(1-\log_{abc}{a})}=1-\log_{abc}{a}\quad\quad\cdots(1)$

By symmetry (or similarly), we have other two equations for $y$ and $z$. Finally adding all those three equations, we get

$\frac{x}{1+x}+\frac{y}{1+y}+\frac{z}{1+z}==3-\log_{abc}{a}-\log_{abc}{b}-\log_{abc}{c}=3-\log_{abc}{abc}=2$

Note that the restrictions appear in the first line.

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