# The ratio of two positive real numbers is p+sqrt(p^2-q^2):p-sqrt(p^2-q^2)then find their ratio of AM and GM?

Jun 11, 2018

$\frac{p}{q}$.

#### Explanation:

Let the nos. be $x \mathmr{and} y , \text{ where, x,y} \in {\mathbb{R}}^{+}$.

By what is given, $x : y = \left(p + \sqrt{{p}^{2} - {q}^{2}}\right) : \left(p - \sqrt{{p}^{2} - {q}^{2}}\right)$.

$\therefore \frac{x}{p + \sqrt{{p}^{2} - {q}^{2}}} = \frac{y}{p - \sqrt{{p}^{2} - {q}^{2}}} = \lambda , \text{ say}$.

$\therefore x = \lambda \left(p + \sqrt{{p}^{2} - {q}^{2}}\right) \mathmr{and} y = \lambda \left(p - \sqrt{{p}^{2} - {q}^{2}}\right)$.

Now, the AM $A$ of $x , y$ is, $A = \frac{x + y}{2} = \lambda p$, and, their

GM $G = \sqrt{x y} = \sqrt{{\lambda}^{2} \left\{{p}^{2} - \left({p}^{2} - {q}^{2}\right)\right\}} = \lambda q$.

Clearly, $\text{the desired ratio} = \frac{A}{G} = \frac{\lambda p}{\lambda q} = \frac{p}{q}$.

Jun 11, 2018

$\frac{p}{q}$

#### Explanation:

I am going to use the same notation as in this answer . In fact there is no real necessity of this solution (as the problem has already been solved quite nicely) - except that it illustrates the use of a technique I love very much!

According to the problem

$\frac{x}{y} = \frac{p + \sqrt{{p}^{2} - {q}^{2}}}{p - \sqrt{{p}^{2} - {q}^{2}}}$

Using componendo and dividendo (this is the favorite technique I alluded to above) we get

$\frac{x + y}{x - y} = \frac{p}{\sqrt{{p}^{2} - {q}^{2}}} \implies$

${\left(\frac{x + y}{x - y}\right)}^{2} = {p}^{2} / \left({p}^{2} - {q}^{2}\right) \implies$

${\left(x + y\right)}^{2} / \left({\left(x + y\right)}^{2} - {\left(x - y\right)}^{2}\right) = {p}^{2} / \left({p}^{2} - \left({p}^{2} - {q}^{2}\right)\right) \implies$

${\left(x + y\right)}^{2} / \left(4 x y\right) = {p}^{2} / {q}^{2} \implies$

$\frac{x + y}{2 \sqrt{x y}} = \frac{p}{q}$

• which is the required AM : GM ratio.