In a triangle, if `r_(1) gt r_(2) gt r_(3),` then show `a gtb gt c.`

Answer

To prove that if \( r_1 > r_2 > r_3 \) in a triangle, then \( a > b > c \), we can follow these steps:

### Step 1: Understand the relationship between the inradii and sides of the triangle
Given that \( r_1, r_2, r_3 \) are the radii of the excircles opposite to the sides \( a, b, c \) respectively, we have the following formulas:
- \( r_1 = \frac{\Delta}{s-a} \)
- \( r_2 = \frac{\Delta}{s-b} \)
- \( r_3 = \frac{\Delta}{s-c} \)

where \( \Delta \) is the area of the triangle and \( s \) is the semi-perimeter given by \( s = \frac{a+b+c}{2} \).

### Step 2: Set up the inequalities
From the given condition \( r_1 > r_2 > r_3 \), we can write:
\[ \frac{\Delta}{s-a} > \frac{\Delta}{s-b} > \frac{\Delta}{s-c} \]

### Step 3: Cancel out the common term
Since \( \Delta > 0 \) for a valid triangle, we can safely cancel \( \Delta \) from all parts of the inequality:
\[ \frac{1}{s-a} > \frac{1}{s-b} > \frac{1}{s-c} \]

### Step 4: Take the reciprocal of the inequalities
Taking the reciprocal of the inequalities reverses the order:
\[ s-a < s-b < s-c \]

### Step 5: Rearranging the inequalities
Now, we can rearrange these inequalities to express them in terms of \( a, b, c \):
\[ -a < -b < -c \]
This implies:
\[ a > b > c \]

### Conclusion
Thus, we have shown that if \( r_1 > r_2 > r_3 \), then \( a > b > c \).