Find the largest number of real numbers x1,...,xn such that for all i≠j,
|xi−xj|>1/100(1+xixj).
To find the largest number of real numbers x1, x2, ..., xn that satisfy the given condition, we can start by analyzing the inequality.
According to the given condition, for any two distinct indices i and j, the absolute difference between xi and xj, denoted as |xi - xj|, must be greater than 1/100(1 + xi*xj).
Let's consider a simple case with two numbers x1 and x2. The inequality becomes:
|x1 - x2| > 1/100(1 + x1*x2)
To proceed, we can assume without loss of generality that x1 is less than x2. This implies (x1 - x2) is negative, so we can remove the absolute value.
-(x1 - x2) > 1/100(1 + x1*x2)
Multiplying both sides by -100 to remove the fraction:
100(x2 - x1) > 1 + x1*x2
Simplifying:
100x2 - 100x1 > 1 + x1*x2
Rearranging the terms:
x1*x2 - 100x1 - 100x2 + 1 < 0
Now, let's complete the square for this quadratic inequality. Adding and subtracting (100/2)^2 = 2500:
x1*x2 - 100x1 - 100x2 + 2501 - 2500 < 0
Factoring the quadratic:
(x1 - 100)(x2 - 100) - 2500 < 0
At this point, we have obtained our inequality in factored form. To satisfy the inequality, we need the expression (x1 - 100)(x2 - 100) - 2500 to be negative. This means that one of the factors must be positive and the other must be negative.
To maximize the number of real solutions, let's consider the case when all numbers (x1, x2, ..., xn) are distinct. In this case, we would have (n-1) negative factors and 1 positive factor.
Since the quadratic expression (x1 - 100)(x2 - 100) - 2500 does not change sign when x1 or x2 is positive/negative (as they are multiplied), we can conclude that there can only be at most one positive factor.
Therefore, the largest number of real numbers (x1, x2, ..., xn) that satisfy the given inequality is two.
In summary, we found that the largest number of real numbers x1, x2, ..., xn satisfying the given inequality is two, given that all the numbers are distinct.