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Eyad

  • 3 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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  1. Eyad
    • 3 years ago
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    A.s=Arithmetic sequence .

  2. Eyad
    • 3 years ago
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    @experimentX :can u help plz :)

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

  4. Eyad
    • 3 years ago
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    yea ,thats right

  5. experimentX
    • 3 years ago
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    \[ 4c^2 - 2a^2 = 6d^2 - 3b^2 \] \( 4c^2 + 3b^2 = 6d^2 + 2a^2\)

  6. experimentX
    • 3 years ago
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    they all look like geometric mean

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

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

  9. experimentX
    • 3 years ago
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    Arithmetic mean is always greater than Geometric mean

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

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