I am stuck in the following olympiad problem: Suppose that a,b,c,d>=0 and a+b+c+d=4. Prove that a/(1+b^2c)+b/(1+c^2d)+c/(1+d^2a)+d/(1+a^2b)>=2

Kailey Vargas 2022-09-11 Answered
a 1 + b 2 c + b 1 + c 2 d + c 1 + d 2 a + d 1 + a 2 b 2
I am stuck in the following olympiad problem:
Suppose that a , b , c , d 0 and a + b + c + d = 4. Prove that
a 1 + b 2 c + b 1 + c 2 d + c 1 + d 2 a + d 1 + a 2 b 2.
Attempt. I tried to use reversed AM-GM technique.
a 1 + b 2 c a = a b 2 c 1 + b 2 c b 1 + c 2 d b = b c 2 d 1 + c 2 d c 1 + d 2 a c = c d 2 a 1 + d 2 a d 1 + a 2 b d = d a 2 b 1 + a 2 b
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Answers (2)

Mohammed Farley
Answered 2022-09-12 Author has 15 answers
By C-S and AM-GM we obtain:
c y c d 1 + a 2 b = c y c d 2 d + a 2 b d ( a + b + c + d ) 2 c y c ( d + a 2 b d ) =
= 16 4 + ( a b + c d ) ( a d + b c ) 16 4 + ( a b + b c + c d + d a 2 ) 2 =
= 16 4 + ( ( a + c ) ( b + d ) 2 ) 2 16 4 + ( ( a + c + b + d 2 ) 2 2 ) 2 = 2.

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themediamafia73
Answered 2022-09-13 Author has 3 answers
c y c d 1 + a 2 b = c y c d 2 d + a 2 b d
Hence by Titu's lemma
c y c d 2 d + a 2 b d ( a + b + c + d ) 2 c y c ( d + a 2 b d ) =
= 16 4 + ( a b + c d ) ( a d + b c )
Now by AM GM inequality we have
a b + b c + c d + a d 2 ( a b + c d ) ( a d + b c )
Hence we get
( a b + b c + c d + a d 2 ) 2 ( a b + c d ) ( a d + b c )
16 4 + ( a b + c d ) ( a d + b c ) 16 4 + ( a b + b c + c d + a d 2 ) 2
Now again by AM GM inequality
a + b + c + d 2 ( a + c ) ( d + b )
16 4 + ( ( a + c ) ( b + d ) 2 ) 2 16 4 + ( ( a + b + c + d 2 ) 2 2 ) 2
= 2.

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