Question

If A is the arithmetic mean and G the geometric mean of two unequal positive real numbers p and q then which of the following is true
1. \(A > G > \frac{G^2}{A}\)
2. \( G > \frac{G^2}{A} > A \)
3. \( \frac{G^2}{A} > A >G \)
4. None of the above

Answer 1

Correct Answer - Option 1 : \(A > G > \frac{G^2}{A}\)

Concept:

The arithmetic mean of two unequal positive real numbers p and q = \(\frac{p+q}{2}\)

The geometric mean of two unequal positive real numbers p and q = √pq

Calculation:

Let A.M > GM

\(\frac{p+q}{2} > \sqrt{pq}\)

Multiply both sides by 2 and square both sides (since both sides are positive numbers, squaring both sides does not change the equality).

⇒ p² + 2pq + q² > 4pq

Subtracting 4pq from both sides:

⇒ p2 - 2pq + q2 > 0

⇒ (p - q)² > 0 (which is always true since a real number squared is always greater than or equal to 0)

Hence AM > GM or A > G     ---- (i)

Let G > G²/A

\(\sqrt{pq}>\frac{(\sqrt{pq})^2}{\frac{p +q}{2}}\)

⇒ \(\frac{p+q}{2} > \sqrt{pq}\)

After reciprocal of both sides.

\(\frac{1}{\frac{p+q}{2}} < \frac{1}{\sqrt{pq}}\) (change in the inequality sign because of reciprocality)

Multiply both sides by p.q

\(\frac{pq}{\frac{p+q}{2}} < \frac{pq}{\sqrt{pq}}\)

or, \(\frac{\sqrt{pq}^2}{\frac{p+q}{2}} < \sqrt{pq}\)

Hence proved, G > G2/A     ---- (ii)

∴ \(A > G > \frac{G^2}{A}\)