Eyad
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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Eyad
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A.s=Arithmetic sequence .

Eyad
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@experimentX :can u help plz :)

experimentX
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\[ 3b^2+4c^2 > 2^{1/2} (2ac+3bd)\]

Eyad
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yea ,thats right

experimentX
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\[ 4c^2  2a^2 = 6d^2  3b^2 \]
\( 4c^2 + 3b^2 = 6d^2 + 2a^2\)

experimentX
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they all look like geometric mean

experimentX
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\( 3b^2 > \sqrt{2a^2 4 c^2 }\)
do similar for other .. you will get answer

Eyad
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wait a min ,from where u got the last relation which is 3b^2 <(2a^24c^2)^1/2

experimentX
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Arithmetic mean is always greater than Geometric mean

Eyad
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OH,k i got it ..I knew that rule before but never worked on it ,TY :D