If x, y, z > 0, xy ? 1, yz ? 1, zx ? 1 and xyz ? 1, then log (xyz) 1 log (xyz) 1 log (xyz) 1 xy yz zx ? ?

To tackle the question, we need to analyze the relationships between the variables x, y, and z, given the conditions that x, y, z are all greater than zero, and the products xy, yz, zx, and xyz are all greater than one. We will also explore the implications of these conditions on the logarithmic expressions provided.

Understanding the Conditions

We start with the inequalities:

• xy > 1
• yz > 1
• zx > 1
• xyz > 1

Since x, y, and z are all positive, we can take the logarithm of these inequalities. The logarithm function is monotonically increasing, meaning if a > b, then log(a) > log(b).

Applying Logarithms

Taking the natural logarithm (or any logarithm) of each inequality gives us:

• log(xy) > 0
• log(yz) > 0
• log(zx) > 0
• log(xyz) > 0

Now, we can express these logarithms in terms of their components:

• log(xy) = log(x) + log(y)
• log(yz) = log(y) + log(z)
• log(zx) = log(z) + log(x)
• log(xyz) = log(x) + log(y) + log(z)

Summing the Logarithmic Inequalities

Next, we can sum the inequalities:

• log(x) + log(y) > 0
• log(y) + log(z) > 0
• log(z) + log(x) > 0

From these, we can derive that:

• log(x) + log(y) + log(z) > 0

This is because if each pair of logs is positive, the sum of all three must also be positive. This confirms that log(xyz) > 0, which is consistent with our earlier finding that xyz > 1.

Combining the Results

Now, let's analyze the expression you provided:

log(xyz) 1 log(xyz) 1 log(xyz) 1 xy yz zx ? ?

It seems there might be a formatting issue in the expression. However, if we interpret it as needing to show that:

• log(xyz) + log(xyz) + log(xyz) > log(xy) + log(yz) + log(zx)

We can rewrite this as:

• 3 * log(xyz) > log(xy) + log(yz) + log(zx)

Substituting the logarithmic identities we derived earlier, we can see that:

• 3 * (log(x) + log(y) + log(z)) > (log(x) + log(y)) + (log(y) + log(z)) + (log(z) + log(x))

This simplifies to:

• 3 * (log(x) + log(y) + log(z)) > 2 * (log(x) + log(y) + log(z))

Since 3 > 2, this inequality holds true, confirming that the original expression is valid under the given conditions.

Final Thoughts

In summary, the conditions provided lead us to conclude that the logarithmic relationships hold true, reinforcing the idea that if the products of the variables are greater than one, then their logarithmic sums will also reflect that positivity. This is a powerful aspect of logarithmic functions, especially when dealing with inequalities in positive domains.