SOLUTION: By three method prove that : |A

SOLUTION: By three method prove that : |A - B |≥ |A|-|B|

Question 1177437: By three method prove that : |A - B |≥ |A|-|B|
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
**Method 1: Using the Triangle Inequality**
The Triangle Inequality states that for any vectors **A** and **B**:
```
|A + B| ≤ |A| + |B|
```
Let's manipulate this to prove our inequality.
1. **Start with A - B:** We can write A as (A - B) + B.
2. **Apply the Triangle Inequality:**
```
|A| = |(A - B) + B| ≤ |A - B| + |B|
```
3. **Rearrange:** ```
|A| - |B| ≤ |A - B|
```
4. **Since |A - B| = |B - A|, we can also write:**
```
|B| - |A| ≤ |B - A| = |A - B|
```
5. **Combine the inequalities:** Since |A - B| is greater than or equal to both |A| - |B| and |B| - |A|, it must be greater than or equal to the absolute value of the difference:
```
|A - B| ≥ ||A| - |B||
```
6. **Finally, since ||A| - |B|| ≥ |A| - |B|, we have:**
```
|A - B| ≥ |A| - |B|
```
**Method 2: Using Properties of Absolute Value**
1. **Recall that |x|² = x² for any real number x.** Hence,
```
|A - B|^2 = (A - B) \cdot (A - B) = A \cdot A - 2 A \cdot B + B \cdot B = |A|^2 - 2 A \cdot B + |B|^2.
```
2. **By the Cauchy-Schwarz Inequality,**
```
|A \cdot B| \le |A| |B|,
```
so
```
-2 A \cdot B \ge -2 |A| |B|.
```
3. **Then**
```
|A - B|^2 = |A|^2 - 2 A \cdot B + |B|^2 \ge |A|^2 - 2 |A| |B| + |B|^2 = (|A| - |B|)^2.
```
4. **Taking square roots of both sides,** we get
```
|A - B| \ge ||A| - |B|| \ge |A| - |B|.
```
**Method 3: Geometric Interpretation**
1. **Vectors as Geometric Objects:** Consider **A** and **B** as vectors in space. Then **A**, **B**, and **A - B** form the sides of a triangle.
2. **Triangle Inequality:** The Triangle Inequality states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. In our case, this means:
```
|A| + |B| > |A - B|
```
3. **Rearrange:**
```
|A| > |A - B| - |B|
```
4. **Since |A - B| = |B - A|, we can also get**
```
|B| > |A - B| - |A|,
```
which we can write as
```
|A| - |A - B| < |B|.
```
5. **Since |A| - |B| is between |A - B| - |B| and |A - B| - |A|,**
```
|A - B| \ge ||A| - |B|| \ge |A| - |B|.
```
These three methods demonstrate the validity of the inequality |A - B| ≥ |A| - |B|.
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