Any ideas for this cyclic inequality? If a,b,c,d>0 and a+b+c+d=8, prove that a/(b^2+2b)+ b/(c^2+2c)+c/(d^2+2d)+\d/(a^2+2a)>=(16)/((a+c)(b+d)).

Frankie Burnett 2022-11-21 Answered
Any ideas for this cyclic inequality?
If a , b , c , d > 0 and a + b + c + d = 8, prove that
a b 2 + 2 b + b c 2 + 2 c + c d 2 + 2 d + d a 2 + 2 a 16 ( a + c ) ( b + d ) .
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Answers (1)

Zoey Benitez
Answered 2022-11-22 Author has 18 answers
By Holder and AM-GM we obtain:
c y c a b 2 + 2 b = c y c a b c y c a ( b + 2 ) c y c a b ( b + 2 ) c y c a b c y c a ( b + 2 ) ( a + b + c + d ) 3 ( a + c ) ( b + d ) c y c ( a b + 2 a ) =
= 512 ( a + c ) ( b + d ) ( ( a + c ) ( b + d ) + 16 ) 512 ( a + c ) ( b + d ) ( ( a + c + b + d 2 ) 2 + 16 ) =
= 16 ( a + c ) ( b + d ) .
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