We are tasked with proving the inequality:

\[
a^a \cdot b^b \cdot c^c > abc^{\frac{a+b+c}{3}}
\]

### Step-by-step solution:

#### Step 1: Restate the inequality

The inequality can be written as:

\[
a^a \cdot b^b \cdot c^c > abc^{\frac{a+b+c}{3}}
\]

Taking the natural logarithm of both sides, we aim to simplify the inequality. Applying the logarithmic property \(\log(xy) = \log(x) + \log(y)\), we get:

\[
\log(a^a \cdot b^b \cdot c^c) > \log\left(abc^{\frac{a+b+c}{3}}\right)
\]

This simplifies to:

\[
a \log a + b \log b + c \log c > \log a + \log b + \log c + \frac{a+b+c}{3} \log c
\]

#### Step 2: Compare powers and simplify

To proceed,

Prove that a^a•b^b•c^c > abc^(a+b+c)/3

Explore the step-by-step proof of the inequality a^a•b^b•c^c > abc^(a+b+c)/3. This advanced-level problem involves the use of logarithmic properties and the Arithmetic Mean-Geometric Mean Inequality. Suitable for advanced high school or undergraduate students studying inequalities and exponential functions.

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