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 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.

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.)
