Question

If g is the geometric mean between two distinct positive numbers a and b then show that 1/g-a+1/g-b=1/g

Answer 1

If g is the geometric mean between two distinct positive numbers a and b then show that $\frac{1}{g-a}+\frac{1}{g-b}=\frac{1}{g}$

a/c to question, g is the geometric mean between two distinct positive numbers a and b.

so, g = $\sqrt{ab}$.....(1)

now, LHS = $\frac{1}{g-a}+\frac{1}{g-b}$

from equation (1),

$=\frac{1}{\sqrt{ab}-a}+\frac{1}{\sqrt{ab}-b}\\\\=\frac{1}{\sqrt{a}(\sqrt{b}-\sqrt{a})}+\frac{1}{\sqrt{b}(\sqrt{a}-\sqrt{b})}\\\\=\frac{1}{(\sqrt{b}-\sqrt{a})}\left[\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right]\\\\=\frac{1}{(\sqrt{b}-\sqrt{a})}\left[\frac{(\sqrt{b}-\sqrt{a})}{\sqrt{ab}}\right]\\\\=\frac{1}{\sqrt{ab}}=\frac{1}{g}=RHS$

hence proved