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

Explanation

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

### Step 1: Set up the inequality
We want to compare \( \sqrt{11} - \sqrt{6} \) and \( \sqrt{17} - \sqrt{12} \). We can set up the inequality:
\[ \sqrt{11} - \sqrt{6} \quad ? \quad \sqrt{17} - \sqrt{12} \]

### Step 2: Rearranging the inequality
Let's assume that \( \sqrt{17} - \sqrt{12} \) is greater than \( \sqrt{11} - \sqrt{6} \). This gives us:
\[ \sqrt{11} + \sqrt{12} > \sqrt{17} + \sqrt{6} \]

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

### Step 4: Expand both sides
Now we expand both sides using the formula \( (a + b)^2 = a^2 + b^2 + 2ab \):
\[ 11 + 12 + 2\sqrt{11 \cdot 12} > 17 + 6 + 2\sqrt{17 \cdot 6} \]
This simplifies to:
\[ 23 + 2\sqrt{132} > 23 + 2\sqrt{102} \]

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

### Step 6: Square both sides again
Now we square both sides again:
\[ 132 > 102 \]
This statement is true.

### Conclusion
Since our assumption that \( \sqrt{17} - \sqrt{12} \) is greater than \( \sqrt{11} - \sqrt{6} \) led to a true statement, we conclude that:
\[ \sqrt{11} - \sqrt{6} > \sqrt{17} - \sqrt{12} \]

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

---