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


(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.

  • $\begingroup$ Have you found an example of such a triple? $\endgroup$
    – lulu
    Aug 26, 2022 at 12:07
  • $\begingroup$ Such triples are easy to construct...and then you'll know what the answer is. Of course, you'll still have to show that the pattern you found holds generally but knowing the answer is a big assist. $\endgroup$
    – lulu
    Aug 26, 2022 at 12:13
  • 2
    $\begingroup$ Here is a question for you to think about later. (I am not asking you to answer it here.) Why did you do all that algebra? It kind of looks like you are doing random calculations with no particular aim in mind. If that's the case then you are unlikely to find anything useful. To solve a problem, you need a plan. $\endgroup$
    – David
    Aug 26, 2022 at 12:21
  • $\begingroup$ @David +1 : Very nice comment. $\endgroup$ Aug 26, 2022 at 13:28

3 Answers 3


$a^2, b^2, c^2$ 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.


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

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

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


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.)


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