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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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A.s=Arithmetic sequence .
@experimentX :can u help plz :)
\[ 3b^2+4c^2 > 2^{1/2} (2ac+3bd)\]

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Other answers:

yea ,thats right
\[ 4c^2 - 2a^2 = 6d^2 - 3b^2 \] \( 4c^2 + 3b^2 = 6d^2 + 2a^2\)
they all look like geometric mean
\( 3b^2 > \sqrt{2a^2 4 c^2 }\) do similar for other .. you will get answer
wait a min ,from where u got the last relation which is 3b^2 <(2a^24c^2)^1/2
Arithmetic mean is always greater than Geometric mean
OH,k i got it ..I knew that rule before but never worked on it ,TY :D

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