x,y,z>0 and 1/(1+x)+1/(1+y)+1/(1+z)=2.
prove that xyz<=1/8
To prove that xyz ≤ 1/8 given the condition 1/(1+x) + 1/(1+y) + 1/(1+z) = 2, we can use the method of contradiction.
Assume that xyz > 1/8, and let's start by solving the equation 1/(1+x) + 1/(1+y) + 1/(1+z) = 2 for better understanding.
First, let's find a common denominator to simplify the equation:
1/(1+x) + 1/(1+y) + 1/(1+z) = 2
(1+y)(1+z) + (1+x)(1+z) + (1+x)(1+y) = 2(1+x)(1+y)(1+z)
Expanding each term, we get:
(1+y+z+yz) + (1+x+z+xz) + (1+x+y+xy) = 2(1+x+y+z+xy+xz+yz+xyz)
3 + 2x + 2y + 2z + xy + xz + yz = 2 + 2x + 2y + 2z + 2xy + 2xz + 2yz + 2xyz
Canceling out similar terms on both sides, we are left with:
1 + xy + xz + yz = 2xyz
Now, let's substitute this back into the original equation:
1 + xy + xz + yz = 2xyz
Since xyz > 1/8, we can rewrite it as:
1 + xy + xz + yz > 2xyz > 1/8
Now, let's simplify further:
8 + 8xy + 8xz + 8yz > 16xyz > 1
Dividing through by 16 on all sides:
1/2 + xy/2 + xz/2 + yz/2 > xyz/2 > 1/16
Since x, y, z are positive, we know that xy, xz, yz > 0.
Therefore,
1/2 + xy/2 + xz/2 + yz/2 > 1/2
This contradicts our assumption that 1/(1+x) + 1/(1+y) + 1/(1+z) = 2, as the left side is greater than the right side.
Hence, our assumption that xyz > 1/8 is false. Consequently, we can conclude that xyz ≤ 1/8.