Problem 1556

1. Applying the Power-Mean Inequality:
By the Power-Mean Inequality, we have:
$$\sqrt[5]{\frac{a^5 + b^5}{2}} \geq \frac{a + b}{2}$$
Raising both sides to the power of 5, we get:
$$\frac{a^5 + b^5}{2} \geq \left(\frac{a + b}{2}\right)^5$$

2. Simplifying the Right-Hand Side:
We know that:
$$\left(\frac{a + b}{2}\right)^5 = \left(\frac{a + b}{2}\right)^4 \cdot \frac{a + b}{2}$$
Using the AM-GM inequality, we have:
$$\left(\frac{a + b}{2}\right)^4 \geq \left(\sqrt[4]{ab}\right)^4 = ab$$
Therefore:
$$\left(\frac{a + b}{2}\right)^5 \geq ab \cdot \frac{a + b}{2}$$

3. Combining the Inequalities:
From the above, we get:
$$\frac{a^5 + b^5}{2} \geq ab \cdot \frac{a + b}{2}$$
Multiplying both sides by 2, we obtain:
$$a^5 + b^5 \geq ab(a + b)$$

4. Substituting into the Original Expression:
We need to prove:
$$\frac{ab}{ab + a^5 + b^5} + \frac{bc}{bc + b^5 + c^5} + \frac{ca}{ca + c^5 + a^5} \leq 1$$
Using the inequality $a^5 + b^5 \geq ab(a + b)$, we have:
$$\frac{ab}{ab + a^5 + b^5} \leq \frac{ab}{ab + ab(a + b)} = \frac{ab}{ab(1 + a + b)} = \frac{1}{1 + a + b}$$

5. Summing the Cyclic Terms:
Similarly, we get:
$$\frac{bc}{bc + b^5 + c^5} \leq \frac{1}{1 + b + c}$$
$$\frac{ca}{ca + c^5 + a^5} \leq \frac{1}{1 + c + a}$$
Summing these inequalities, we have:
$$\frac{1}{1 + a + b} + \frac{1}{1 + b + c} + \frac{1}{1 + c + a} \leq 1$$

6. **Using the Condition $abc = 1$:**
Since $abc = 1$, we can use the fact that the sum of the reciprocals of the terms in the cyclic sum is less than or equal to 1:
$$\frac{1}{1 + a + b} + \frac{1}{1 + b + c} + \frac{1}{1 + c + a} \leq 1$$

Therefore, we have shown that:
$$\frac{ab}{ab + a^5 + b^5} + \frac{bc}{bc + b^5 + c^5} + \frac{ca}{ca + c^5 + a^5} \leq 1$$

$\blacksquare$