Which is greater `sqrt(11)-sqrt(6) or sqrt(17)-sqrt(12)` ?

Explanation

To determine which is greater between \( \sqrt{11} - \sqrt{6} \) and \( \sqrt{17} - \sqrt{12} \), we can follow these steps:

### Step 1: Set up the inequality
Assume that:
\[ \sqrt{17} - \sqrt{12} > \sqrt{11} - \sqrt{6} \]

### Step 2: Rearrange the inequality
Rearranging gives:
\[ \sqrt{17} - \sqrt{12} - \sqrt{11} + \sqrt{6} > 0 \]
This can be rewritten as:
\[ \sqrt{17} + \sqrt{6} > \sqrt{11} + \sqrt{12} \]

### Step 3: Square both sides
Since both sides are positive, we can square both sides without changing the inequality:
\[ (\sqrt{17} + \sqrt{6})^2 > (\sqrt{11} + \sqrt{12})^2 \]

### Step 4: Expand both sides
Expanding both sides:
\[ 17 + 6 + 2\sqrt{17 \cdot 6} > 11 + 12 + 2\sqrt{11 \cdot 12} \]
This simplifies to:
\[ 23 + 2\sqrt{102} > 23 + 2\sqrt{132} \]

### Step 5: Simplify the inequality
Subtract 23 from both sides:
\[ 2\sqrt{102} > 2\sqrt{132} \]
Dividing both sides by 2 gives:
\[ \sqrt{102} > \sqrt{132} \]

### Step 6: Square both sides again
Squaring both sides yields:
\[ 102 > 132 \]
This is a contradiction, as 102 is not greater than 132.

### Conclusion
Since our assumption that \( \sqrt{17} - \sqrt{12} > \sqrt{11} - \sqrt{6} \) led to a contradiction, it must be false. Therefore, we conclude that:
\[ \sqrt{11} - \sqrt{6} > \sqrt{17} - \sqrt{12} \]

### Final Answer
Thus, \( \sqrt{11} - \sqrt{6} \) is greater than \( \sqrt{17} - \sqrt{12} \).

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