Question 1210209: If a,b,c are in Harmonic Progression, show that 1/a + 1/(b+c) , 1/b + 1/(c+a) , 1/c + 1/(a+b) are also in Harmonic Progression
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Since $a, b, c$ are in Harmonic Progression (HP), their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in Arithmetic Progression (AP).
This means that the difference between consecutive terms is constant:
$\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}$
$\frac{a-b}{ab} = \frac{b-c}{bc}$
$bc(a-b) = ab(b-c)$
Dividing by $abc$ (assuming $a, b, c \neq 0$), we get:
$\frac{a-b}{a} = \frac{b-c}{c}$
$c(a-b) = a(b-c)$
$ca - cb = ab - ac$
$2ac = ab + bc$
Now, let's consider the terms $\frac{1}{a} + \frac{1}{b+c}$, $\frac{1}{b} + \frac{1}{c+a}$, $\frac{1}{c} + \frac{1}{a+b}$.
For these terms to be in Harmonic Progression, their reciprocals must be in Arithmetic Progression. Let's find the reciprocals:
$x = \frac{1}{\frac{1}{a} + \frac{1}{b+c}} = \frac{a(b+c)}{b+c+a}$
$y = \frac{1}{\frac{1}{b} + \frac{1}{c+a}} = \frac{b(c+a)}{c+a+b}$
$z = \frac{1}{\frac{1}{c} + \frac{1}{a+b}} = \frac{c(a+b)}{a+b+c}$
For $x, y, z$ to be in AP, we need $y - x = z - y$, or $2y = x + z$.
$2 \left( \frac{b(c+a)}{a+b+c} \right) = \frac{a(b+c)}{a+b+c} + \frac{c(a+b)}{a+b+c}$
Since the denominators are the same, we can equate the numerators:
$2b(c+a) = a(b+c) + c(a+b)$
$2bc + 2ab = ab + ac + ca + cb$
$2bc + 2ab = ab + 2ac + cb$
$bc + ab = 2ac$
This is the same condition we derived from the fact that $a, b, c$ are in HP.
Therefore, the reciprocals of $\frac{1}{a} + \frac{1}{b+c}$, $\frac{1}{b} + \frac{1}{c+a}$, $\frac{1}{c} + \frac{1}{a+b}$ are in Arithmetic Progression, which means that $\frac{1}{a} + \frac{1}{b+c}$, $\frac{1}{b} + \frac{1}{c+a}$, $\frac{1}{c} + \frac{1}{a+b}$ are in Harmonic Progression.
Final Answer: The final answer is $\boxed{\frac{1}{a} + \frac{1}{b+c} , \frac{1}{b} + \frac{1}{c+a} , \frac{1}{c} + \frac{1}{a+b} \text{ are in Harmonic Progression}}$
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