Question

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

Answer 1

A.s=Arithmetic sequence .

Answer 2

@experimentX :can u help plz :)

Answer 3

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

Answer 4

yea ,thats right

Answer 5

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

Answer 6

they all look like geometric mean

Answer 7

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

Answer 8

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

Answer 9

Arithmetic mean is always greater than Geometric mean

Answer 10

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