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anonymous
 4 years ago
If(2a^2 ,3b^2 ,4c^2 ,6d^2) in A.s ,,Prove That :
3b^2+4c^2 > 2^1/2 (2ac+3bd)
anonymous
 4 years ago
If(2a^2 ,3b^2 ,4c^2 ,6d^2) in A.s ,,Prove That : 3b^2+4c^2 > 2^1/2 (2ac+3bd)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0A.s=Arithmetic sequence .

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@experimentX :can u help plz :)

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1\[ 3b^2+4c^2 > 2^{1/2} (2ac+3bd)\]

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1\[ 4c^2  2a^2 = 6d^2  3b^2 \] \( 4c^2 + 3b^2 = 6d^2 + 2a^2\)

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1they all look like geometric mean

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1\( 3b^2 > \sqrt{2a^2 4 c^2 }\) do similar for other .. you will get answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait a min ,from where u got the last relation which is 3b^2 <(2a^24c^2)^1/2

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1Arithmetic mean is always greater than Geometric mean

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OH,k i got it ..I knew that rule before but never worked on it ,TY :D
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