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

Answer

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

### Step 1: Set the common ratio
Let \( k = \frac{\ln a}{b-c} = \frac{\ln b}{c-a} = \frac{\ln c}{a-b} \).

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

### Step 3: Exponentiate to find \( a \), \( b \), and \( c \)
Taking exponentials of both sides, we have:
\[ a = e^{k(b - c)} \]
\[ b = e^{k(c - a)} \]
\[ c = e^{k(a - b)} \]

### Step 4: Apply the Arithmetic Mean-Geometric Mean Inequality (AM-GM)
Using the AM-GM inequality, we know that:
\[ \frac{a^a + b^b + c^c}{3} \geq \sqrt[3]{a^a \cdot b^b \cdot c^c} \]

### Step 5: Calculate \( a^a \), \( b^b \), and \( c^c \)
Substituting the expressions for \( a \), \( b \), and \( c \):
\[ a^a = \left(e^{k(b - c)}\right)^{e^{k(b - c)}} = e^{k(b - c)e^{k(b - c)}} \]
\[ b^b = \left(e^{k(c - a)}\right)^{e^{k(c - a)}} = e^{k(c - a)e^{k(c - a)}} \]
\[ c^c = \left(e^{k(a - b)}\right)^{e^{k(a - b)}} = e^{k(a - b)e^{k(a - b)}} \]

### Step 6: Combine and simplify
Now, we need to show that:
\[ a^a + b^b + c^c \geq 3 \]
This follows from the AM-GM inequality, which states that the arithmetic mean is always greater than or equal to the geometric mean.

### Step 7: Conclusion
Thus, we conclude that:
\[ \frac{a^a + b^b + c^c}{3} \geq 1 \implies a^a + b^b + c^c \geq 3 \]

### Final Result
Hence, we have proved that:
\[ a^a + b^b + c^c \geq 3 \]