If a1's are all +ve real numbers then prove that:
(1+a1+a12)(1+a2+a22).........((1+an+an2)≥3na1a2..........an
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2 Answers
Lokesh Verma
·2009-01-18 22:20:50
apply AM GM on 1, ai, ai2
(1+a1+a12)/3>=(1.a1.a12)1/3
thus,
(1+a1+a12)>=3a1
now multiply this for all ai's
thus the final result :)