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 , 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: we first square both sides:
Step 2: Expand the left-hand side
Expand the left side: So, the inequality becomes:
Step 3: Consider two cases based on the sign of
Case 1:
- In this case, .
- The inequality becomes: Subtract from both sides: Divide by (which flips the inequality): This simplifies to: This is true because implies .
Case 2:
- In this case, .
- The inequality becomes: Subtract from both sides: Divide by : This simplifies to: This is true because implies .
Conclusion
In both cases, the inequality holds. Therefore, we have proved that:
Would you like more details on any part of this proof? Here are some related questions you might find interesting:
- How does the triangle inequality apply to square roots?
- Can we extend this inequality to higher powers, such as and ?
- How does this inequality relate to the concept of the metric in geometry?
- What is the geometric interpretation of the inequality ?
- 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
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