If $(b + c) , (c + a) , (a + b)$ are in H.P. then find the relation between $\dfrac{a}{b+c} , \dfrac{b}{c+a} , \dfrac{c}{a+b}$ .

Question

If $(b + c) , (c + a) , (a + b)$ are in H.P. then $\dfrac{a}{b+c} , \dfrac{b}{c+a} , \dfrac{c}{a+b}$ are in $(i)$ A.P. $(ii)$ G.P. $(iii)$ H.P. $(iv)$ None of These.

What I Tried:- I have that $(b+c),(c+a),(a+b)$ are in H.P. .

$\implies \dfrac{1}{b+c}, \dfrac{1}{c+a}, \dfrac{1}{a+b}$ are in A.P. .

$\implies \dfrac{1}{c+a} - \dfrac{1}{b+c} = \dfrac{1}{a+b} - \dfrac{1}{c+a}.$

$\implies \dfrac{(b + c) - (c + a)}{(c + a)(b+c)} = \dfrac{(c+a) - (a+b)}{(a+b)(c+a)}.$

$\implies \dfrac{(b - a)}{(b + c)} = \dfrac{(c - b)}{(b + a)}.$

$\implies (b^2 - a^2) = (c^2 - b^2).$

$\implies 2b^2 = a^2 + c^2.$

So I was able to prove that $a^2 , b^2 , c^2$ are in A.P. , but I am not sure how to show the claim in the question.

Can anyone help me? Thank You.

Answer 1

As , $a^2,b^2,c^2$ & $\frac{1}{(b+c)}, \frac{1}{(c+a)}, \frac{1}{(a+b)}$ are in $AP$ then , $$\frac{2}{(c+a)}=\frac{1}{(b+c)}+\frac{1}{(a+b)}$$ $$\frac{2b}{(c+a)}=\frac{b}{(b+c)}+\frac{b}{(a+b)}$$ $$\frac{2b}{(c+a)}=\frac{ba+\color{green}{b^2+b^2}+bc}{(a+b)(b+c)}$$

Since we have $a^2,b^2,c^2$ are in $AP$ , we have : $$2b^2=a^2+c^2$$

So , finally $$\frac{2b}{(c+a)}=\frac{\color{blue}{a^2+ab}+\color{red}{c^2+cb}}{(a+b)(b+c)}$$ $$\frac{2b}{(c+a)}=\frac{\color{blue}{a(a+b)}+\color{red}{c(b+c)}}{\color{blue}{(a+b)}\color{red}{(b+c)}}$$ $$\color{green}{\frac{2b}{(c+a)}=\frac{a}{(b+c)}+\frac{c}{(a+b)}}$$

We're DONE . Hence its in $AP$ . Please consider UpVoting

Answer 2

Hint 1:

Add $1$ to each term as mentioned in the comments by @SathvikAcharya

Hint 2:

If $p,q,r$ is an arithmetic progression, $\lambda p,\lambda q,\lambda r$ is also an arithmetic progression where $\lambda$ is a real number

Answer 3

Just to show another way this might be written:

Once you demonstrated that $ \ \dfrac{1}{b+c} \ , \ \dfrac{1}{c+a} \ , \ \dfrac{1}{a+b} \ $ in arithmetic progression implies that

$$ \ \dfrac{1}{c+a} - \dfrac{1}{b+c} \ = \ \dfrac{1}{a+b} - \dfrac{1}{c+a} \ \implies \ \dfrac{(b - a)}{(c + a)·(b+c)} \ = \ \dfrac{(c - b)}{(a+b)·(c+a)} \ \ , \quad \quad \mathbf{[1]} $$
it is already becoming clear that $ \ \dfrac{a}{b+c} \ , \ \dfrac{b}{c+a} \ , \ \dfrac{c}{a+b} \ $ is unlikely to be a geometric series as there are differences of terms in the numerators of the ratios in equation $ \ \mathbf{[1]} \ \ . $ Similarly, the "single-number" numerators of the ratios in the sequence do not appear consistent with their reciprocals being in arithmetic progression. So we will want to test whether the ratios themselves are in arithmetic progression; otherwise, we will need to choose $ \ \mathbf{(iv)} \ \ . \ \ $ (This looks to be a reasonable route of investigation, since the differences of sequence terms will produce the same products of factors in the denominators.)

$$ \dfrac{b}{c+a} \ - \ \dfrac{a}{b+c} \ \ = \ \ \dfrac{ b^2 \ + \ bc \ - \ ac \ - \ a^2}{(c + a)·(b+c)} \ \ = \ \ \dfrac{ (b - a) \ · \ [ \ (b + a) \ + c \ ] }{(c + a)·(b+c)} \ \ , $$ $$ \dfrac{c}{a+b} \ - \ \dfrac{b}{c+a} \ \ = \ \ \dfrac{ c^2 \ + \ ac \ - \ ab \ - \ b^2}{(a + b)·(c+a)} \ \ = \ \ \dfrac{ (c - b) \ · \ [ \ (c + b) \ + a \ ] }{(a + b)·(c+a)} \ \ . $$

These two ratios have the common factor $ \ a + b + c \ $ in their numerators, so by equation $ \ \mathbf{[1]} \ \ , $ we can establish that $ \ \dfrac{ (b - a) \ · \ ( \ a + b + c \ ) }{(c + a)·(b+c)} \ = \ \dfrac{ (c - b) \ · \ ( \ a + b + c \ ) }{(a + b)·(c+a)} \ \ . $ Hence, there is a common difference between the terms of the sequence $ \ \dfrac{a}{b+c} \ , \ \dfrac{b}{c+a} \ , \ \dfrac{c}{a+b} \ \ , $ making this an arithmetic progression [choice $ \ \mathbf{(i)} \ ] \ . $