AM-HM

If the ratio of AM-to-HM for two positive numbers $a$ and $b$ is $\frac{m}{n}$ , then we can write:

$\frac{a + b}{2} \div \frac{2}{a^-1 + b^-1} = \frac{a + b}{2} \div \frac{2ab}{a + b} = \frac{a + b}{2} \cdot \frac{a + b}{2ab} = \frac{(a + b)^2}{4ab} = \frac{m}{n}.$

After expanding and simplifying, we now obtain:

$\frac{(a + b)^2}{4ab} = \frac{m}{n} \Rightarrow \frac{a^2 + 2ab + b^2}{4ab} = \frac{m}{n} \Rightarrow \frac{a}{b} + \frac{b}{a} + 2 = 4*\frac{m}{n} = \frac{a}{b} + \frac{b}{a} = 4*\frac{m}{n} - 2$

and if we substitute $u = \frac{a}{b}$ , we now get $u + \frac{1}{u} = \frac{4m}{n} - 2 \Rightarrow u^2 - (\frac{4m}{n} - 2)u + 1 = 0$ .

Using the Quadratic Formula (and keeping in mind $u > 0$ for $m,n > 0$ ), we next find:

$u = \frac{(4 \cdot \frac{m}{n} -2) + \sqrt{( 4 \cdot \frac{m}{n} -2)^2 - 4(1)(1)}}{2} = \frac{(4 \cdot \frac{m}{n} -2) + \sqrt{( 16 \cdot \frac{m^2}{n^2} - 16 \cdot \frac{m}{n} + 4 - 4}}{2} = \frac{ \frac{4m - 2n}{n} + \sqrt{(\frac{16m^2 - 16mn}{n^2})}}{2} = \frac{(4m - 2n) + 4 \cdot \sqrt{(m^2 - mn)}}{2n} = \frac{(2m - n) + 2 \cdot \sqrt{(m^2 - mn)}}{n}$ .

Now taking note that the numerator equals: $(2m - n) + 2\sqrt{m^2 - mn} = m + (m-n) + 2\sqrt{m(m-n)} = (\sqrt{m})^2 + 2 \cdot \sqrt{m} \cdot \sqrt{m-n} + (\sqrt{m-n})^2 = (\sqrt{m} + \sqrt{m-n})^2$

and the denominator equals: $n = m - (m-n) = (\sqrt{m})^2 - (\sqrt{m-n})^2 = (\sqrt{m} - \sqrt{m-n})(\sqrt{m} + \sqrt{m-n})$

we now ultimately obtain: $u = \frac{a}{b} = \frac{(\sqrt{m} + \sqrt{m-n})^2}{(\sqrt{m} - \sqrt{m-n})(\sqrt{m} + \sqrt{m-n})} = \boxed{\frac{\sqrt{m} + \sqrt{m-n}}{\sqrt{m} - \sqrt{m-n}}}$ .

NCERT problem