If $a,b,c$ be in Harmonic Progression, prove that $\dfrac{1}{a}+\dfrac{1}{b+c}, \dfrac{1}{b}+\dfrac{1}{c+a}, \dfrac{1}{c}+\dfrac{1}{a+b}$ are also in Harmonic Progression.
$\dfrac{1}{a}+\dfrac{1}{b+c}, \dfrac{1}{b}+\dfrac{1}{c+a}, \dfrac{1}{c}+\dfrac{1}{a+b}$ can be rewritten as $\dfrac{a+b+c}{ab+ac},\dfrac{a+b+c}{ab+bc},\dfrac{a+b+c}{ac+bc}$.
The reciprocal must be in Arithmetical Progression:
$\dfrac{ab+ac}{a+b+c},\dfrac{ab+bc}{a+b+c},\dfrac{ac+bc}{a+b+c}$
Since denominator is the same, it can be disregarded:
$a(b+c),b(a+c),c(a+b)$ are in Arithmetical Progression.
Since $a,b,c$ are in Harmonic Progression, $\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}$ are in Arithmetical Progression. Thus $\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{2}{b}$ or $\dfrac{a+c}{ac}=\dfrac{2}{b} \Rightarrow 2ac=b(a+c)$
hence $a(b+c), 2ac, c(a+b)$ are in Arithmetical Progression. Check with Arithmetical Mean: $ab+ac+ac+bc=4ac \Rightarrow ab+2ac+bc=4ac \Rightarrow b(a+c)=2ac$
Would this proof be correct? Thanks.
If a, b, c in HP: $b = \frac{2ac}{a+c}$ that can be also expressed as $\frac{b}{a}+\frac{b}{c} =2$
$\frac{\frac{1}{b}+\frac{1}{c+a}}{\frac{1}{a}+\frac{1}{b+c}} + \frac{\frac{1}{b}+\frac{1}{c+a}}{\frac{1}{c}+\frac{1}{a+b}} = \frac{a(b+c)}{b(c+a)} + \frac{c(a+b)}{b(c+a)} = \frac{b(c+a)+2ac}{b(c+a)} = 1 + \frac{2ac}{b(c+a)} = 2$.