Prove that a b + 2 c + 3 d </m

sebadillab0 2022-07-03 Answered
Prove that a b + 2 c + 3 d + b c + 2 d + 3 a + c d + 2 a + 3 b + d a + 2 b + 3 c 2 3 .
I was thinking of trying a b c since the inequality is cyclic. I am not sure how to use this, though, or if this would help simplify the inequality. It also doesn't look like I can really use AM-GM or Cauchy-Schwarz.
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Answers (2)

lofoptiformfp
Answered 2022-07-04 Author has 16 answers
We need only note that
  a b + 2 c + 3 d + b c + 2 d + 3 a + c d + 2 a + 3 b + d a + 2 b + 3 c =   a 2 a b + 2 a c + 3 a d + b 2 b c + 2 b d + 3 a b + c 2 c d + 2 a c + 3 b c + d 2 a d + 2 b d + 3 c d   ( a + b + c + d ) 2 4 ( a b + a c + a d + b c + b d + c d ) =   a 2 + b 2 + c 2 + d 2 + 2 ( a b + a c + a d + b c + b d + c d ) 4 ( a b + a c + a d + b c + b d + c d )   2 3 ( a b + a c + a d + b c + b d + c d ) + 2 ( a b + a c + a d + b c + b d + c d ) 4 ( a b + a c + a d + b c + b d + c d ) =   2 3 .
EDIT:
( a + b + c + d ) 2 = ( a a b + 2 a c + 3 a d a b + 2 a c + 3 a d + b b c + 2 b d + 3 a b b c + 2 b d + 3 a b + c c d + 2 a c + 3 b c c d + 2 a c + 3 b c + d a d + 2 b d + 3 c d a d + 2 b d + 3 c d ) 2 ( a 2 a b + 2 a c + 3 a d + b 2 b c + 2 b d + 3 a b + c 2 c d + 2 a c + 3 b c + d 2 a d + 2 b d + 3 c d ) [ 4 ( a b + a c + a d + b c + b d + c d ) ] .
That is,
  a 2 a b + 2 a c + 3 a d + b 2 b c + 2 b d + 3 a b + c 2 c d + 2 a c + 3 b c + d 2 a d + 2 b d + 3 c d   ( a + b + c + d ) 2 4 ( a b + a c + a d + b c + b d + c d ) .
Moreover,
( a + b + c + d ) 2 = a 2 + b 2 + c 2 + d 2 + 2 ( a b + a c + a d + b c + b d + c d ) = 1 3 [ ( a 2 + b 2 ) + ( a 2 + c 2 ) + ( a 2 + d 2 ) + ( b 2 + c 2 ) + ( b 2 + d 2 ) ) + ( c 2 + d 2 ) ] + 2 ( a b + a c + a d + b c + b d + c d ) 2 3 ( a b + a c + a d + b c + b d + c d ) + 2 ( a b + a c + a d + b c + b d + c d ) = 8 3 ( a b + a c + a d + b c + b d + c d ) .
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spockmonkey40
Answered 2022-07-05 Author has 4 answers
By C-S c y c a b + 2 c + 3 d ( a + b + c + d ) 2 c y c ( 4 a b + 2 a c ) 2 3 , where the last inequality it's
s y m ( a b ) 2 0. Done!
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