Problem 1464

To find the greatest possible value of
$$E(a,b,c) = \frac{a^n}{a^{2n+1} + b^{2n}c + bc^{2n}} + \frac{b^n}{b^{2n+1} + c^{2n}a + ca^{2n}} + \frac{c^n}{c^{2n+1} + a^{2n}b + ab^{2n}},$$
we start by using the given condition:
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1.$$

1. Applying the AM-GM Inequality:
By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq 3 \sqrt[3]{\frac{1}{abc}}.$$
Given that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$, it follows that:
$$1 \geq 3 \sqrt[3]{\frac{1}{abc}} \implies \frac{1}{3} \geq \sqrt[3]{\frac{1}{abc}} \implies 27 \geq abc.$$

2. Using Cauchy-Schwarz Inequality:
We apply the Cauchy-Schwarz inequality in the form:
$$(a^{2n+1} + b^{2n}c + bc^{2n}) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \geq (a^n + b^n + c^n)^2.$$
Given $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$, we have:
$$a^{2n+1} + b^{2n}c + bc^{2n} \geq (a^n + b^n + c^n)^2.$$

3. **Analyzing the Expression $E(a,b,c)$:**
We need to maximize:
$$E(a,b,c) = \frac{a^n}{a^{2n+1} + b^{2n}c + bc^{2n}} + \frac{b^n}{b^{2n+1} + c^{2n}a + ca^{2n}} + \frac{c^n}{c^{2n+1} + a^{2n}b + ab^{2n}}.$$
Using the inequality derived from Cauchy-Schwarz:
$$\frac{a^n}{a^{2n+1} + b^{2n}c + bc^{2n}} \leq \frac{a^n}{(a^n + b^n + c^n)^2}.$$
Similarly for the other terms, we get:
$$E(a,b,c) \leq \frac{a^n}{(a^n + b^n + c^n)^2} + \frac{b^n}{(a^n + b^n + c^n)^2} + \frac{c^n}{(a^n + b^n + c^n)^2}.$$
Factoring out the common denominator:
$$E(a,b,c) \leq \frac{a^n + b^n + c^n}{(a^n + b^n + c^n)^2} = \frac{1}{a^n + b^n + c^n}.$$

4. Maximizing the Expression:
To maximize $E(a,b,c)$, we need to minimize $a^n + b^n + c^n$. Given $a = b = c = 3$ (since $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$ and $a, b, c$ are positive numbers), we have:
$$a^n + b^n + c^n = 3^n + 3^n + 3^n = 3 \cdot 3^n = 3^{n+1}.$$
Therefore:
$$E(a,b,c) \leq \frac{1}{3^{n+1}}.$$

5. Conclusion:
The maximum value of $E(a,b,c)$ is achieved when $a = b = c = 3$, giving:
$$E(a,b,c) = \frac{1}{3^{n+1}}.$$

The final answer is $\boxed{\frac{1}{3^{n+1}}}$.