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
To prove the inequality, let's start by simplifying the expressions:
sqrt(x + b) - sqrt(x - a) ≥ sqrt(x + a) - sqrt(x - b)
First, let's square both sides of the inequality to remove the square roots:
(sqrt(x + b) - sqrt(x - a))^2 ≥ (sqrt(x + a) - sqrt(x - b))^2
Expanding both sides:
(x + b) - 2sqrt[(x + b)(x - a)] + (x - a) ≥ (x + a) - 2sqrt[(x + a)(x - b)] + (x - b)
Next, let's cancel out the common terms:
2b - 2sqrt[(x + b)(x - a)] ≥ 2a - 2sqrt[(x + a)(x - b)]
Dividing both sides by 2:
b - sqrt[(x + b)(x - a)] ≥ a - sqrt[(x + a)(x - b)]
Finally, to complete the proof, we need to show that the left-hand side is greater than or equal to the right-hand side:
b ≥ a - sqrt[(x + a)(x - b)] + sqrt[(x + b)(x - a)]
Now, let's simplify the right-hand side:
b ≥ a - sqrt[(x + a)(x - b)] + sqrt[(x + b)(x - a)]
b ≥ a - sqrt(x^2 - bx + ax - ab) + sqrt(x^2 + bx - ax - ab)
b ≥ a - sqrt(x^2 - ab + (a - b)x) + sqrt(x^2 - ab + (b - a)x)
Since a ≥ b ≥ 0, we can see that a - b ≥ 0 and b - a ≤ 0. Therefore, (a - b)x ≥ 0 and (b - a)x ≤ 0. This allows us to make the following simplification:
b ≥ a - sqrt(x^2 - ab) + sqrt(x^2 - ab)
Now, let's combine the two square roots and simplify further:
b ≥ a + sqrt(x^2 - ab) - sqrt(x^2 - ab)
Both square roots cancel out:
b ≥ a
Since we are given that x ≥ a ≥ b ≥ 0, the original inequality:
sqrt(x + b) - sqrt(x - a) ≥ sqrt(x + a) - sqrt(x - b)
is true.
Hence, we have shown that for any real numbers x, a, b where x ≥ a ≥ b ≥ 0, the inequality sqrt(x + b) - sqrt(x - a) ≥ sqrt(x + a) - sqrt(x - b) holds.