If `(Ina)/(b-c)=(Inb)/(c-a)=(Inc)/(a-b)`, prove the following .
`a+b+cge3`

Answer

To prove that \( a + b + c \geq 3 \) given the condition \( \frac{\ln a}{b-c} = \frac{\ln b}{c-a} = \frac{\ln c}{a-b} \), we can follow these steps:

### Step 1: Set the common ratio
Let \( k \) be the common value of the ratios:
\[ \frac{\ln a}{b-c} = \frac{\ln b}{c-a} = \frac{\ln c}{a-b} = k \]

### Step 2: Express \( \ln a \), \( \ln b \), and \( \ln c \)
From the above equations, we can express \( \ln a \), \( \ln b \), and \( \ln c \) in terms of \( k \):
\[ \ln a = k(b - c) \]
\[ \ln b = k(c - a) \]
\[ \ln c = k(a - b) \]

### Step 3: Exponentiate to express \( a \), \( b \), and \( c \)
Now, we can exponentiate these equations to express \( a \), \( b \), and \( c \):
\[ a = e^{k(b - c)} \]
\[ b = e^{k(c - a)} \]
\[ c = e^{k(a - b)} \]

### Step 4: Use the Arithmetic Mean-Geometric Mean Inequality
We apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[ \frac{a + b + c}{3} \geq \sqrt[3]{abc} \]
This implies:
\[ a + b + c \geq 3 \sqrt[3]{abc} \]

### Step 5: Calculate \( abc \)
Now, substituting the expressions for \( a \), \( b \), and \( c \):
\[ abc = e^{k(b - c)} \cdot e^{k(c - a)} \cdot e^{k(a - b)} \]
This simplifies to:
\[ abc = e^{k((b - c) + (c - a) + (a - b))} = e^{k \cdot 0} = e^0 = 1 \]

### Step 6: Substitute back into AM-GM
Now substituting \( abc = 1 \) into the AM-GM inequality:
\[ a + b + c \geq 3 \sqrt[3]{1} = 3 \]

### Conclusion
Thus, we have shown that:
\[ a + b + c \geq 3 \]