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Inequalities Involving Mathematical Means

Arithmetic me...

Question

Arithmetic mean of two numbers is $a$ and their geometric mean is $b$ . Show that these numbers are $a + \sqrt{a^{2} - b^{2}}$ and $a - \sqrt{a^{2} - b^{2}}$ .

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Solution

Let the two number by 'x' & 'y'
Given $\frac{x + y}{2} = a$ - (i)
$\sqrt{x y} = b$
$x y = b^{2}$ - (ii)
$(x - y)^{2} = (x + y)^{2} - 4 x y$
$= 4 a^{2} - 4 b^{2}$
$∴ x - y = 2 \sqrt{a^{2} - b^{2}}$
$x + y = 2 a$
$∴ 2 x = 2 a + 2 \sqrt{a^{2} - b^{2}}$
$∴ x = a + \sqrt{a^{2} - b^{2}}$
$∴ y = a - \sqrt{a^{2} - b^{2}}$

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