Math Problem Statement

knowing that sqrt(a+b)<=sqrt(a)+sqrt(b) , prove that abs(sqrt(a)-sqrt(b))<=sqrt(abs(a-b))

Solution

To prove that abab|\sqrt{a} - \sqrt{b}| \leq \sqrt{|a - b|}, we'll start by squaring both sides of the inequality and then show that the inequality holds.

Step 1: Start with the squared version of the inequality

Given the inequality: abab|\sqrt{a} - \sqrt{b}| \leq \sqrt{|a - b|} we first square both sides: (ab)2ab(\sqrt{a} - \sqrt{b})^2 \leq |a - b|

Step 2: Expand the left-hand side

Expand the left side: (ab)2=a2ab+b(\sqrt{a} - \sqrt{b})^2 = a - 2\sqrt{a}\sqrt{b} + b So, the inequality becomes: a2ab+baba - 2\sqrt{a}\sqrt{b} + b \leq |a - b|

Step 3: Consider two cases based on the sign of aba - b

Case 1: aba \geq b

  • In this case, ab=ab|a - b| = a - b.
  • The inequality becomes: a2ab+baba - 2\sqrt{a}\sqrt{b} + b \leq a - b Subtract a+ba + b from both sides: 2ab2b-2\sqrt{a}\sqrt{b} \leq -2b Divide by 2-2 (which flips the inequality): abb\sqrt{a}\sqrt{b} \geq b This simplifies to: ab\sqrt{a} \geq \sqrt{b} This is true because aba \geq b implies ab\sqrt{a} \geq \sqrt{b}.

Case 2: a<ba < b

  • In this case, ab=ba|a - b| = b - a.
  • The inequality becomes: a2ab+bbaa - 2\sqrt{a}\sqrt{b} + b \leq b - a Subtract b+ab + a from both sides: 2ab2a-2\sqrt{a}\sqrt{b} \leq -2a Divide by 2-2: aba\sqrt{a}\sqrt{b} \geq a This simplifies to: ba\sqrt{b} \geq \sqrt{a} This is true because bab \geq a implies ba\sqrt{b} \geq \sqrt{a}.

Conclusion

In both cases, the inequality holds. Therefore, we have proved that: abab|\sqrt{a} - \sqrt{b}| \leq \sqrt{|a - b|}

Would you like more details on any part of this proof? Here are some related questions you might find interesting:

  1. How does the triangle inequality apply to square roots?
  2. Can we extend this inequality to higher powers, such as a1na^{\frac{1}{n}} and b1nb^{\frac{1}{n}}?
  3. How does this inequality relate to the concept of the metric in geometry?
  4. What is the geometric interpretation of the inequality a+ba+b\sqrt{a+b} \leq \sqrt{a} + \sqrt{b}?
  5. How does this proof change if we use complex numbers instead of real numbers?

Tip: When proving inequalities, consider squaring both sides to simplify expressions, but be mindful of maintaining the correct inequality direction.

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Math Problem Analysis

Mathematical Concepts

Inequalities
Square Roots
Absolute Values

Formulas

|√a - √b| ≤ √|a - b|
(√a - √b)^2 = a - 2√a√b + b

Theorems

Triangle Inequality
Properties of Absolute Values

Suitable Grade Level

Grades 10-12