given real numbers x, a, b with x>= a>= b>= 0, show that sqrt x+b - sqrt x-a >= sqrt x+a - sqrt x-b
sqrt x+b - sqrt x-a >= sqrt x+a - sqrt x-b
√(x+b)-√(x-a) >= √(x+a)-√(x-b)
since x>=a>=b, we have
x+b >= 2b
x-a >= 0
x+a >= 2a >= 2b
x-b >= x-a
see what you can do with that
To show that sqrt(x+b) - sqrt(x-a) >= sqrt(x+a) - sqrt(x-b), where x >= a >= b >= 0, we can use a proof by contradiction.
Assume the opposite, that is, suppose sqrt(x+b) - sqrt(x-a) < sqrt(x+a) - sqrt(x-b).
Squaring both sides of the inequality, we get:
(x + b) - 2 * sqrt((x + b) * (x - a)) + (x - a) < (x + a) - 2 * sqrt((x + a) * (x - b)) + (x - b)
Expanding and simplifying, we have:
2 * sqrt((x + b) * (x - a)) - 2 * sqrt((x + a) * (x - b)) < 2 * (a - b)
Dividing by 2, we get:
sqrt((x + b) * (x - a)) - sqrt((x + a) * (x - b)) < a - b
Now, let's consider the values under the square roots. Since x >= a >= b >= 0, both (x + b) and (x - a) are non-negative. Similarly, (x + a) and (x - b) are non-negative. This means that all terms under the square roots are non-negative, and hence, the square roots themselves are real numbers.
Taking the square of both sides of the inequality, we have:
((x + b) * (x - a)) + ((x + a) * (x - b)) - 2 * sqrt((x + b) * (x - a)) * sqrt((x + a) * (x - b)) < (a - b)^2
Expanding and simplifying, we get:
2x^2 + 2ab - 2 * (x^2 + ab) - 2 * sqrt((x + b) * (x - a)) * sqrt((x + a) * (x - b)) < (a - b)^2
This further simplifies to:
2 * sqrt((x + b) * (x - a)) * sqrt((x + a) * (x - b)) > (a - b)^2
Notice that the left side of the inequality is positive since both sqrt((x + b) * (x - a)) and sqrt((x + a) * (x - b)) are non-negative. The right side of the inequality is also positive since a >= b >= 0.
Therefore, we have a contradiction. Our initial assumption was false, and therefore, sqrt(x+b) - sqrt(x-a) >= sqrt(x+a) - sqrt(x-b) for x >= a >= b >= 0.