If `log_(a) bc =x, log_(b) ca=y, log_( c)ab =z`, prove that `1/(x+1) + 1/(y+1) + 1/(z+1)=1`

Answer

To prove that

\[ \frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} = 1 \]

given that

\[ \log_a(bc) = x, \quad \log_b(ca) = y, \quad \log_c(ab) = z, \]

we will start by rewriting the logarithmic expressions in terms of their properties.

### Step 1: Rewrite the logarithmic expressions
Using the change of base formula, we can express \(x\), \(y\), and \(z\) as follows:

\[ x = \log_a(bc) = \log_a b + \log_a c \]
\[ y = \log_b(ca) = \log_b c + \log_b a \]
\[ z = \log_c(ab) = \log_c a + \log_c b \]

### Step 2: Express \(x + 1\), \(y + 1\), and \(z + 1\)
Now, we can express \(x + 1\), \(y + 1\), and \(z + 1\):

\[ x + 1 = \log_a b + \log_a c + 1 = \log_a b + \log_a c + \log_a a = \log_a(bac) \]
\[ y + 1 = \log_b c + \log_b a + 1 = \log_b c + \log_b a + \log_b b = \log_b(cab) \]
\[ z + 1 = \log_c a + \log_c b + 1 = \log_c a + \log_c b + \log_c c = \log_c(abc) \]

### Step 3: Take the reciprocals
Now, we take the reciprocals:

\[ \frac{1}{x + 1} = \frac{1}{\log_a(bac)}, \quad \frac{1}{y + 1} = \frac{1}{\log_b(cab)}, \quad \frac{1}{z + 1} = \frac{1}{\log_c(abc)} \]

### Step 4: Use the change of base formula
Using the change of base formula, we can rewrite these as:

\[ \frac{1}{\log_a(bac)} = \log_{bac} a, \quad \frac{1}{\log_b(cab)} = \log_{cab} b, \quad \frac{1}{\log_c(abc)} = \log_{abc} c \]

### Step 5: Combine the terms
Now we combine these terms:

\[ \log_{bac} a + \log_{cab} b + \log_{abc} c \]

### Step 6: Apply the logarithmic identity
Using the identity that states \(\log_a b + \log_b c + \log_c a = 1\) for any positive \(a\), \(b\), and \(c\):

\[ \log_{bac} a + \log_{cab} b + \log_{abc} c = 1 \]

### Conclusion
Thus, we have shown that:

\[ \frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} = 1 \]