If a, b, c are in A.P., show that 1/(√b +√c) , 1/(√c + √a), 1/(√a + √b) are in A.P.

Question

If a, b, c are in A.P., show that \(\frac{1}{\sqrt{b} +\sqrt{c}}\), \(\frac{1}{\sqrt{c} +\sqrt{a}}\), \(\frac{1}{\sqrt{a} +\sqrt{b}}\) are in A.P.

Answer 1

Given 

a, b, c are in A.P. ⇒ 2b = a + c

Now ,

  \(\frac{1}{\sqrt{b} +\sqrt{c}}\), \(\frac{1}{\sqrt{c} +\sqrt{a}}\), \(\frac{1}{\sqrt{a} +\sqrt{b}}\) are in A.P., if 

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

⇒ \(\frac{2}{\sqrt{c} +\sqrt{a}}\) = \(\frac{1}{\sqrt{b} +\sqrt{c}}\)+ \(\frac{1}{\sqrt{a} +\sqrt{b}}\)

⇒ \(\frac{2}{\sqrt{c} +\sqrt{a}}\) = \(\frac{\sqrt{a} + \sqrt{b} + \sqrt{b} + \sqrt{c}}{(\sqrt{b} + \sqrt{c}) (\sqrt{a} + \sqrt{b})}\)

⇒ 2(√b + √c)(√a +√b) = (√c +√a)(√a +2√b +√c)

⇒ 2(√ab + b + √ac + √bc) = (√ac + 2√bc + c + a + 2√ba + √ac)

⇒ 2√ab +2b +2√ac +2√bc = 2√ac + 2√bc + c + a + 2√ba

⇒ 2b = a + c, which is true as a, b, c are in A.P.

⇒  \(\frac{1}{\sqrt{b} +\sqrt{c}}\), \(\frac{1}{\sqrt{c} +\sqrt{a}}\), \(\frac{1}{\sqrt{a} +\sqrt{b}}\) are in A.P