If a^2, b^2, c^2 are in A.P., prove that a b+c , b c+a , c a+b are in A.P.

a b+c , b c+a , c a+b are in A.P if b a+ c - a b+ c = c a+ b - b a+ c LHS= b a+ c - a b+ c b^2+bc-a^2-ac (a+c)(b+c) (b-a)(a+b+c) (a+c)(b+c) .....(1) RHS= a a+b = b a+c ca+c^2-b^2-ab (a+b)(b+c) (c-d)(a+b+c) (a+b)(b+c) .....(2) and a^2, b^2, c^2 are in A.P ^2-a^2=c^2-b^2 .....(3) Substituting b^2-a^2 with c^2-b^2 (1)=(2) a b+c , b a+c , c a+b are in A.P

If a^2, b^2, c^2 are in A.P., prove that a b+c , b c+a , c a+b ar…

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Question

If $\text{a}^2,\ \text{b}^2,\ \text{c}^2$ are in A.P., prove that $\frac{\text{a}}{\text{b}+\text{c}},\frac{\text{b}}{\text{c}+\text{a}},\frac{\text{c}}{\text{a}+\text{b}}$ are in A.P.

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$\frac{\text{a}}{\text{b}+\text{c}},\frac{\text{b}}{\text{c}+\text{a}},\frac{\text{c}}{\text{a}+\text{b}}$ are in A.P if $\frac{\text{b}}{\text{a}+\text{c}}-\frac{\text{a}}{\text{b}+\text{c}}=\frac{\text{c}}{\text{a}+\text{b}}-\frac{\text{b}}{\text{a}+\text{c}}$
$\text{LHS}=\frac{\text{b}}{\text{a}+\text{c}}-\frac{\text{a}}{\text{b}+\text{c}}$
$\Rightarrow\frac{\text{b}^2+\text{bc}-\text{a}^2-\text{ac}}{(\text{a}+\text{c})(\text{b}+\text{c})}$
$\Rightarrow\frac{(\text{b}-\text{a})(\text{a}+\text{b}+\text{c})}{(\text{a}+\text{c})(\text{b}+\text{c})}\ .....(1)$
$\text{RHS}=\frac{\text{a}}{\text{a}+\text{b}}=\frac{\text{b}}{\text{a}+\text{c}}$
$\Rightarrow\frac{\text{ca}+\text{c}^2-\text{b}^2-\text{ab}}{(\text{a}+\text{b})(\text{b}+\text{c})}$
$\Rightarrow\frac{(\text{c}-\text{d})(\text{a}+\text{b}+\text{c})}{(\text{a}+\text{b})(\text{b}+\text{c})}\ .....(2)$
and
$\text{a}^2,\ \text{b}^2,\ \text{c}^2$ are in A.P
$\therefore\text{b}^2-\text{a}^2=\text{c}^2-\text{b}^2\ .....(3)$
Substituting $\text{b}^2-\text{a}^2$ with $\text{c}^2-\text{b}^2$
$(1)=(2)$
$\therefore\frac{\text{a}}{\text{b}+\text{c}},\ \frac{\text{b}}{\text{a}+\text{c}},\ \frac{\text{c}}{\text{a}+\text{b}}$ are in A.P

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