If A,G,H are respectively the A.M., G.M. and H.M. of three +ive numbers p,q,r then prove that
(x - p) (x - q) (x - r) = $x^{3} - 3 A x^{2} + 3 \frac{G^{3}}{H} x - G^{3} = 0$

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Solution

$A = \frac{p + q + r}{3} = \frac{s_{1}}{3}$
$∴ s_{1} = 3 A$
G $= {p q r}^{1 / 3}$
$∴ G^{3} = p q r = s_{3}$
$\frac{1}{H} = \frac{1}{3} (\frac{1}{p} + \frac{1}{q} + \frac{1}{r}) = \frac{\sum p q}{3 p q r} = \frac{s_{2}}{3 s_{3}} ∴ s_{2} = \frac{3 S_{3}}{H} = \frac{3 G^{3}}{H}$
now $(x - p) (x - q) (x - r) = x^{3} - x^{2} s_{1} + x s_{2} - s_{3}$

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