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

To prove that \( |\sqrt{a} - \sqrt{b}| \leq \sqrt{|a - b|} \), 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:
\[
|\sqrt{a} - \sqrt{b}| \leq \sqrt{|a - b|}
\]
we first square both sides:
\[
(\sqrt{a} - \sqrt{b})^2 \leq |a - b|
\]

### Step 2: Expand the left-hand side
Expand the left side:
\[
(\sqrt{a} - \sqrt{b})^2 = a - 2\sqrt{a}\sqrt{b} + b
\]
So, the inequality becomes:
\[
a - 2\sqrt{a}\sqrt{b} + b \leq |a - b|
\]

### Step 3: Consider two cases based on the sign of \(a - b\)

**Case 1: \(a \geq b\)**
- In this case, \(|a - b| = a - b\).
- The inequality becomes:
\[
a - 2\sqrt{a}\sqrt{b} + b \leq a - b
\]
Subtract \(a + b\) from both sides:
\[
-2\sqrt{a}\sqrt{b} \leq -2b
\]
Divide by \(-2\) (which flips the inequality):
\[
\sqrt{a}\sqrt{b} \geq b
\]
This simplifies to:
\[
\sqrt{a} \geq \sqrt{b}
\]
This is true because \(a \geq b\) implies \(\sqrt{a} \geq \sqrt{b}\).

**Case 2: \(a < b\)**
- In this case, \(|a - b| = b - a\).
- The inequality becomes:
\[
a - 2\sqrt{a}\sqrt{b} + b \leq b - a
\]
Subtract \(b + a\) from both sides:
\[
-2\sqrt{a}\sqrt{b} \leq -2a
\]
Divide by \(-2\):
\[
\sqrt{a}\sqrt{b} \geq a
\]
This simplifies to:
\[
\sqrt{b} \geq \sqrt{a}
\]
This is true because \(b \geq a\) implies \(\sqrt{b} \geq \sqrt{a}\).

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

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 \(a^{\frac{1}{n}}\) and \(b^{\frac{1}{n}}\)?
3. How does this inequality relate to the concept of the metric in geometry?
4. What is the geometric interpretation of the inequality \( \sqrt{a+b} \leq \sqrt{a} + \sqrt{b} \)?
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.

Explore a detailed proof of the inequality |√a - √b| ≤ √|a - b|, applying concepts such as inequalities, square roots, and absolute values. This mathematical solution is suitable for high school students in Grades 10-12 and includes step-by-step expansion and case analysis.

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