Question

If a, b, c are in AP, show that

$\frac{1}{(\sqrt{b}+\sqrt{c})},\frac{1}{(\sqrt{c}+\sqrt{a})},\frac{1}{(\sqrt{a}+\sqrt{b})}$ are in AP.

Answer 1

Since a, b, c are in AP, we have

2b = (a + c). . . . .(i)

Now, $\frac{1}{(\sqrt{b}+\sqrt{c})},\frac{1}{(\sqrt{c}+\sqrt{a})},\frac{1}{(\sqrt{a}+\sqrt{b})}$ will be in AP

if $\frac{1}{(\sqrt{c}+\sqrt{a})}-\frac{1}{(\sqrt{b}+\sqrt{c})}=\frac{1}{(\sqrt{a}+\sqrt{b})}-\frac{1}{(\sqrt{c}+\sqrt{a})}$

i.e., if $\frac{(\sqrt{b}+\sqrt{c})-(\sqrt{c}+\sqrt{a})}{(\sqrt{c}+\sqrt{a})(\sqrt{b}+\sqrt{c})}=\frac{(\sqrt{c}+\sqrt{a})-(\sqrt{a}+\sqrt{b})}{(\sqrt{a}+\sqrt{b})(\sqrt{c}+\sqrt{a})}$

i.e., if $\frac{(\sqrt{b}-\sqrt{a})}{(\sqrt{c}+\sqrt{a})(\sqrt{b}+\sqrt{c})}=\frac{(\sqrt{c}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})(\sqrt{c}+\sqrt{a})}$

i.e., if $\frac{(\sqrt{b}-\sqrt{a})}{(\sqrt{b}+\sqrt{c})}=\frac{(\sqrt{c}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})}$

i.e., if b - a = c - b

i.e., if 2b = a + c, which is true by (i).

$\therefore$ a, b, c are in $AP\Rightarrow \frac{1}{(\sqrt{b}+\sqrt{c})},\frac{1}{(\sqrt{c}+\sqrt{a})},\frac{1}{(\sqrt{a}+\sqrt{b})}$ are in AP.