Given 1+a+b+c=2abc and positivity of real numbers a,b,c, we are asked to prove that (ab)/(1+a+b)+(bc)/(1+b+c)+(ac)/(1+a+c)>=3/2 If d=a+b+c I got as far as to simplify the inequality into 1/a+1/b+1/c+1/(d−a)+1/(d−b)+1/(d−c)>=3 From a=(1+b+c)/(2bc−1) I also can prove that ab+ac+bc>=3/2 But cannot manage to get to the desired result.

Massatfy 2022-08-16 Answered
Prove a b 1 + a + b + b c 1 + b + c + a c 1 + a + c 3 2 for a, b, c positive and 1 + a + b + c = 2 a b c
If d = a + b + c I got as far as to simplify the inequality into
1 a + 1 b + 1 c + 1 d a + 1 d b + 1 d c 3
From a = 1 + b + c 2 b c 1 I also can prove that
a b + a c + b c 3 2
But cannot manage to get to the desired result.
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Answers (1)

Jaydan Gilbert
Answered 2022-08-17 Author has 16 answers
We need to prove next equivalent inequality
c y c ( a b 1 + a + b + 1 ) 9 2
or
c y c ( a + 1 ) ( b + 1 ) ( c + 1 ) ( 1 + a + b ) ( 1 + c ) 9 2
Then by Cauchy-Schwarz inequality
c y c ( a + 1 ) ( b + 1 ) ( c + 1 ) ( 1 + a + b ) ( 1 + c ) 9 c y c ( a + 1 ) c y c ( 1 + a + b ) ( c + 1 ) =
= 9 c y c ( a + 1 ) c y c ( a + 2 a b + 1 + 2 a ) = 9 2
Now, we have done!
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