Question

If a², b², c² are in A.P., prove that $\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}$ are in A.P.

Answer 1

$$\begin{gathered} {\mathrm{a}}^2,{\mathrm{b}}^2,{\mathrm{c}}^2\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ \therefore 2b^2=a^2+c^2 \\ \Rightarrow b^2-a^2=c^2-b^2 \\ \Rightarrow (b+a)(b-a)=(c-b)(c+b) \\ \Rightarrow \frac{b-a}{c+b}=\frac{c-b}{b+a} \\ \Rightarrow \frac{b-a}{(c+a)(c+b)}=\frac{c-b}{(b+a)(c+a)}\left[\mathrm{Multiplying}\mathrm{both}\mathrm{the}\mathrm{sides}\mathrm{by}\frac{1}{c+a}\right] \\ \Rightarrow \frac{1}{c+a}-\frac{1}{b+c}=\frac{1}{a+b}-\frac{1}{c+a} \\ \therefore \text{'}\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ \mathrm{Multiplying}\mathrm{each}\mathrm{term}\mathrm{by}(a+b+c): \\ \frac{a+b+c}{b+c},\frac{a+b+c}{c+a},\frac{a+b+c}{a+b}\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ \mathrm{Thus},\frac{a}{b+c}+1,\frac{b}{c+a}+1,\frac{c}{a+b}+1\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ \mathrm{Hence},\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \end{gathered}$$