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

Massatfy

Open question

2022-08-16

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.

Answer & Explanation

Jaydan Gilbert

Jaydan Gilbert

Beginner2022-08-17Added 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!

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?