Let $a^2, b^2$ and $c^2$ be three distinct numbers in AP. If $ab + bc + ca = 1$ then $(b + c), (c + a)$ and $(a + b)$ are in which series

Question

Let $a^2, b^2$ and $c^2$ be three distinct numbers in AP. If $ab + bc + ca = 1$ then $(b + c), (c + a)$ and $(a + b)$ are in

(1)AP

(2) GP

(3) HP

(4) none of these

My approach is as follow

$2{b^2} = {a^2} + {c^2} \Rightarrow 2{b^2} + {b^2} = {a^2} + {c^2} + {b^2} \Rightarrow 3{b^2} = {a^2} + {c^2} + {b^2}$

${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right)$

${\left( {a + b + c} \right)^2} = 3{b^2} + 2 \Rightarrow {\left( {a + b + c} \right)^2} - {b^2} = 2{b^2} + 2$

$\Rightarrow \left( {a + b + c - b} \right)\left( {a + b + c + b} \right) = 2{b^2} + 2$

$\Rightarrow \left( {a + c} \right)\left( {a + c + 2b} \right) = 2{b^2} + 2 \Rightarrow {\left( {a + c} \right)^2} + 2b\left( {a + c} \right) = 2{b^2} + 2$

$ \Rightarrow {\left( {a + c} \right)^2} + 2\left( {ab + bc} \right) = 2{b^2} + 2 \Rightarrow {\left( {a + c} \right)^2} + 2\left( {1 - ac} \right) = 2{b^2} + 2$

$\Rightarrow {\left( {a + c} \right)^2} - 2ac = 2{b^2} \Rightarrow {\left( {a + c} \right)^2} = 2\left( {{b^2} + ac} \right)$

$ \Rightarrow \frac{{a + c}}{{{b^2} + ac}} = \frac{2}{{\left( {a + c} \right)}}$

Not able to proceed further.

Answer 1

$$b^2-a^2=c^2-b^2=k(\ne0)\text{say}\implies b^2=a^2+k, c^2-a^2=2k$$

$$(c-b)(c+b)=k\iff b+c=\dfrac k{c-b}, \text{similarly } a+b=\dfrac k{b-a}, c-a=\dfrac{2k}{c+a}$$

$$\dfrac1{b+c}+\dfrac1{a+b}=\dfrac{c-b+b-a}k=\dfrac{2k}{k(c+a)}=? $$

Answer 2

$a^2, b^2, c^2$ are in AP.

Then $a^2+1,b^2+1,c^2+1$ are also in AP.

$a^2+ab+bc+ca, b^2+ab+bc+ca, c^2+ab+bc+ca$ are in AP.

$\begin{align}a^2+ab+bc+ca&=a(a+b) +c(b+a)\\&=(a+c) (a+b) \end{align}$

$\begin{align}b^2+ab+bc+ca&=b(b+a) +c(b+a)\\&=(b+c) (a+b) \end{align}$

$\begin{align}c^2+ab+bc+ca&=c(c+a) +b(c+a)\\&=(b+c) (c+a) \end{align}$

Now divide by $(a+b) (b+c) (c+a) $.

Hence $\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in AP.

$(b+c), (c+a), (a+b) $ are in HP.

Answer 3

For multiple chase exercises, well, not my didactic or structural choice of giving exercises, is always a good idea to have a particular example that may rule out immediately some boxes. So let us take $A^2,B^2,C^2$ to be the squares $1,25,49$ in arithmetic progression. Well, $AB+BC+CA=1\cdot 5+5\cdot7+7\cdot1=5+35+7=47$, so we have to "rescale", i.e. to divide $A,B,C,$ by some number (each by $\sqrt{47}$) to obtain $a,b,c$ with $ab+bc+ca=1$.

Then $(b+c)$, $(c+a)$, $(a+b)$ are "rescaled" versions of $(B+C)=5+7=12$, $(C+A)=7+1=8$, and $(A+B)=1+5=6$, and to see if they are in AP, GP, and/or HP is the same equivalent game in the either original or "rescaled" version. So we used the "rescaled" version to test.

$(1)$ Are the numbers $12$, $8$, $6$ in AP? No, $8+ 8\ne 12+6$.

$(2)$ Are the numbers $12$, $8$, $6$ in GP? No, $8\cdot 8\ne 12\cdot6$.

$(3)$ Are the numbers $12$, $8$, $6$ in HP? Equivalently, are the reciprocals in AP? Yes, $\frac 18\cdot \frac 18= \frac 14=\frac 1{12}+\frac 16$.

So we look if this is the case in general, let $\Pi$ be the product $\Pi=(b+c)(c+a)(a+b)$: $$ \begin{aligned} &\frac 1{b+c} +\frac 1{a+b} - \frac2{c+a} \\ &\qquad=\frac 1\Pi \Big(\ (a+b)(a+c) +(c+a)(c+b) -2(b+a)(b+c)\ \Big) \\ &\qquad=\frac 1\Pi \Big(\ (a^2 +(ab+ac+bc)) + (c^2 + (ab+ac+bc)) -2(b^2+(ab+ac+bc))\ \Big) \\ &\qquad=\frac 1\Pi \Big(\ a^2 + c^2 -2b^2\ \Big) \\ &\qquad=0\ . \end{aligned} $$ Yes, $(3)$ is in general valid. (There is no need for the norming $ab+bc+ca=1$ for $(3)$, but ok, the exercise wants to get arguments for the truth / failure of the possible choices under this supplementary condition.)