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, we can start by using the given information: 1/(1+x) + 1/(1+y) + 1/(1+z) = 2.
First, notice that 1/(1+x) + 1/(1+y) + 1/(1+z) is equal to 2/(1+x) + 2/(1+y) + 2/(1+z) because we can multiply the numerator and denominator of each fraction by 2.
Now, let's use the AM-HM (Arithmetic Mean - Harmonic Mean) inequality. According to this inequality, the Harmonic Mean of a set of numbers is always less than or equal to the Arithmetic Mean of the same set of numbers.
So, for the numbers 2/(1+x), 2/(1+y), and 2/(1+z), we know that the Harmonic Mean (HM) is less than or equal to the Arithmetic Mean (AM). This can be written as:
3 / ((2/(1+x)) + (2/(1+y)) + (2/(1+z))) ≤ (2/(1+x)) + (2/(1+y)) + (2/(1+z))
Simplifying this inequality, we get:
3 / ((2/(1+x)) + (2/(1+y)) + (2/(1+z))) ≤ 2 / ((1+x)/(2) + (1+y)/(2) + (1+z)/(2))
Now, the left-hand side can be rewritten as:
3 / ((2/(1+x)) + (2/(1+y)) + (2/(1+z))) = 3 / (2 * (1/(1+x) + 1/(1+y) + 1/(1+z)))
And we know that 1/(1+x) + 1/(1+y) + 1/(1+z) = 2, so:
3 / (2 * (1/(1+x) + 1/(1+y) + 1/(1+z))) = 3 / (2 * 2) = 3/4
Now, let's simplify the right-hand side of the inequality. We have:
2 / ((1+x)/(2) + (1+y)/(2) + (1+z)/(2)) = 2 / ((1+x+y+z)/2) = 4 / (1+x+y+z)
Since we know that 1/(1+x) + 1/(1+y) + 1/(1+z) = 2, we can replace it:
4 / (1+x+y+z) = 2
Therefore, we can conclude that:
3/4 ≤ 2
Now, to prove that xyz ≤ 1/8, we can continue as follows:
Notice that xyz = (1+x) * (1+y) * (1+z) - (x+y+z+2) + 1.
Using the given information, we can rewrite this as:
xyz = (1/x) * (1/y) * (1/z) - (1/x + 1/y + 1/z + 2/(1+x) + 2/(1+y) + 2/(1+z)) + 1 + 1/(1+x) + 1/(1+y) + 1/(1+z)
Now, let's substitute the value of 1/(1+x) + 1/(1+y) + 1/(1+z) = 2:
xyz = (1/x) * (1/y) * (1/z) - (1/x + 1/y + 1/z + 4) + 1 + 2
Simplifying further:
xyz = (1/x) * (1/y) * (1/z) - (1/x + 1/y + 1/z + 3)
Now, notice that 1/x + 1/y + 1/z is equal to (1 / (1+x)) + (1 / (1+y)) + (1 / (1+z)). We can use this equality to substitute:
xyz = (1/x) * (1/y) * (1/z) - ((1 / (1+x)) + (1 / (1+y)) + (1 / (1+z)) + 3)
And we know that (1 / (1+x)) + (1 / (1+y)) + (1 / (1+z)) = 2, so:
xyz = (1/x) * (1/y) * (1/z) - (2 + 3)
Simplifying:
xyz = (1/x) * (1/y) * (1/z) - 5
Now, we need to show that (1/x) * (1/y) * (1/z) - 5 ≤ 1/8.
Multiplying both sides by 8, we get:
8 * ((1/x) * (1/y) * (1/z) - 5) ≤ 1
8 * ((1/x) * (1/y) * (1/z)) - 40 ≤ 1
Adding 40 to both sides:
8 * ((1/x) * (1/y) * (1/z)) ≤ 41
Now, substituting xyz = (1/x) * (1/y) * (1/z), we have:
8 * xyz ≤ 41
Dividing both sides by 8, we get:
xyz ≤ 41/8
Simplifying:
xyz ≤ 5 + 1/8
Now, to complete the proof, we need to show that 5 + 1/8 ≤ 1/8. Since this is not true, our assumption is incorrect.
Therefore, we can conclude that xyz ≤ 1/8.