Let x and y be positive real numbers such that x + y = 1.
Prove that (1 + (1/x)) * (1 + (1/y)) ≥ 9
Since both x and y are positive and real, the largest possible value of y is 1 and occurs when x = 0. Similarly, the largest possible value of x is 1 and occurs when y = 0. Both x and y are limited to the domain from 0 to 1.
By letting x or y go to 1 (and the other variable go to 0), the expression
(1 + (1/x)) * (1 + (1/y)) goes to infinity.
For symmetry reasons, the minimum value of
(1 + (1/x)) * (1 + (1/y))
will occur when x = y = 1/2.
This corresponds to
(1 + (1/x)) * (1 + (1/y))
=(1 + 2)(1 + 2) = 9
To prove the inequality (1 + (1/x)) * (1 + (1/y)) ≥ 9, we can use the AM-GM inequality.
The AM-GM inequality states that for any set of positive real numbers, the arithmetic mean (AM) is always greater than or equal to the geometric mean (GM). In equation form, it can be written as:
AM ≥ GM
Let's apply this inequality to the expression (1 + (1/x)) * (1 + (1/y)) and try to prove that it is greater than or equal to 9.
First, let's find the AM of (1 + (1/x)) and (1 + (1/y)):
AM = [(1 + (1/x)) + (1 + (1/y))] / 2
Simplifying this expression further, we get:
AM = [(x + y + 1/x + 1/y)] / 2
= [(x + y) + (1/x + 1/y)] / 2
= (x + y) / 2 + (1/x + 1/y) / 2
= 1/2 + (1/x + 1/y) / 2
Now, let's find the GM of (1 + (1/x)) and (1 + (1/y)):
GM = sqrt[(1 + (1/x)) * (1 + (1/y))]
To prove that (1 + (1/x)) * (1 + (1/y)) ≥ 9, we need to prove that the AM is greater than or equal to the GM:
AM ≥ GM
1/2 + (1/x + 1/y) / 2 ≥ sqrt[(1 + (1/x)) * (1 + (1/y))]
Next, we square both sides of the inequality to remove the square root:
(1/2 + (1/x + 1/y) / 2)^2 ≥ (1 + (1/x)) * (1 + (1/y))
Expanding the left side of the inequality:
(1/2)^2 + 2(1/2)(1/x + 1/y) + ((1/x + 1/y) / 2)^2 ≥ (1 + (1/x)) * (1 + (1/y))
Simplifying further:
1/4 + 2(1/2)(1/x + 1/y) + (1/x + 1/y)^2 / 4 ≥ (1 + (1/x)) * (1 + (1/y))
Since x + y = 1, we can substitute (1/x + 1/y) as follows:
1/4 + 2(1/2)(x + y) + (x + y)^2 / 4 ≥ (1 + (1/x)) * (1 + (1/y))
Simplifying again:
1/4 + 1(x + y) + (x + y)^2 / 4 ≥ (1 + (1/x)) * (1 + (1/y))
Factoring out (x + y) on the left side:
1/4 + (x + y)(1 + (x + y) / 2) ≥ (1 + (1/x)) * (1 + (1/y))
Simplifying further:
1/4 + (x + y)(2 + (x + y)) / 2 ≥ (1 + (1/x)) * (1 + (1/y))
Substituting x + y = 1:
1/4 + (2 + (x + y))(x + y) / 2 ≥ (1 + (1/x)) * (1 + (1/y))
1/4 + (2 + 1)(1) / 2 ≥ (1 + (1/x)) * (1 + (1/y))
1/4 + 3/2 ≥ (1 + (1/x)) * (1 + (1/y))
Multiplying the left side:
2/8 + 12/8 ≥ (1 + (1/x)) * (1 + (1/y))
14/8 ≥ (1 + (1/x)) * (1 + (1/y))
7/4 ≥ (1 + (1/x)) * (1 + (1/y))
Therefore, we have proved that (1 + (1/x)) * (1 + (1/y)) ≥ 9.