If x and y are positive real numbers satisfying x+y≤xy, what is the smallest value of x+y?
can you do better than 3+2 </= 6 ?
To find the smallest value of x+y, we need to analyze the given inequality and determine its conditions.
The inequality states: x+y ≤ xy
We can rearrange this inequality as: xy - x - y ≥ 0
Now, let's rewrite it in terms of factors: (x-1)(y-1) ≥ 1
Since x and y are positive real numbers, we know that x > 1 and y > 1.
Now, we can proceed to find the smallest possible value for x+y.
To start, let's assume x = 2 and y = 2. In this case, x+y = 4.
However, with these values, the inequality is not satisfied: (2-1)(2-1) = 1 ≥ 1
So, we need to decrease x and y to make the left-hand side of the inequality smaller than 1.
Now, consider x = 1.5 and y = 1.5. In this case, x+y = 3.
Plugging these values into the inequality, we have: (1.5-1)(1.5-1) = 0.5 ≥ 1
We can see that the inequality is still not satisfied.
Let's try x = 1.1 and y = 1.1. Now, x+y = 2.2.
For this case, plugging the values into the inequality leads to: (1.1-1)(1.1-1) = 0.1 ≥ 1
Again, the inequality is not satisfied.
As a general observation, the closer x and y get to 1, the smaller the left-hand side of the inequality becomes, making it increasingly challenging to satisfy the inequality.
However, if we let x = y = 1, then x+y = 1+1 = 2.
Plugging these values into the inequality gives us: (1-1)(1-1) = 0 ≥ 1
This inequality is satisfied for the values x = y = 1, and x+y = 2 is the smallest possible value that satisfies the given conditions.
Thus, the smallest value of x+y is 2.