If A is the A.M and G is the G.M of two unequal positive real numbers p and q,prove that A>G and |G|>G^2/A?

Question

994 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-16T18:06:32

Kindly refer to a Proof in the Explanation.

Given that, #A and G# are the AM and GM of #p, q in RR^+, pneq#.

#:. A=(p+q)/2, and, G=sqrt(pq).#

#:. A-G=(p+q)/2-sqrt(pq)=(p+q-2sqrt(pq))/2,#

#=(sqrtp-sqrtq)^2/2, and, because, pneq, #

# A-G=(sqrtp-sqrtq)^2/2>0.#

#:. A > G.#

Multiplying this inequality by, #G>0,# we have,

#AG > G^2,#

and, now dividing this by #A>0,# we get,

#G>G^2/A.#

Hence, the desired results.