1assuming it to be true, simplifying, we get
3(ab+bc+ac)\le a^2+b^2+c^2+2ab+2bc+2ca
ab+bc+ac\le a^2+b^2+c^2
which is the expansion of
(a-b)^2 +(b-c)^2 + (c-a)^2 \ge 0
which is obviously true
prove this inequality
conditions:plz dont use titu lemma,tchebychef and cauchy-schwatz....
only alowwed inequality's are a.m >g.m >h.m and mean power............ i am imposing these conditions because i proved it using tchebychef's inequality.........is there any proof using the basic a.m >g.m and mean power inequality
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2 Answers
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11By A.M-G.M.....
apply A.M-G.M on these nos, and see what happens......\frac{a^2+b^2}{2}, \frac{b^2+c^2}{2}, \frac{c^2+a^2}{2}
I mean apply separately, then add 'em up........
