Problem 1390

1. Given the condition $xy + yz + zx = x + y + z$, we need to prove the inequality:
$$\frac{1}{x^2 + y + 1} + \frac{1}{y^2 + z + 1} + \frac{1}{z^2 + x + 1} \leq 1$$

2. By the Cauchy-Schwarz inequality, we have:
$$(x^2 + y + 1)(1 + y + z^2) \geq (x + y + z)^2$$
This implies:
$$\frac{1}{x^2 + y + 1} \leq \frac{1 + y + z^2}{(x + y + z)^2}$$

3. Summing up similar inequalities for $y$ and $z$, we get:
$$\frac{1}{x^2 + y + 1} + \frac{1}{y^2 + z + 1} + \frac{1}{z^2 + x + 1} \leq \frac{1 + y + z^2}{(x + y + z)^2} + \frac{1 + z + x^2}{(x + y + z)^2} + \frac{1 + x + y^2}{(x + y + z)^2}$$

4. Simplifying the right-hand side, we obtain:
$$\frac{1 + y + z^2 + 1 + z + x^2 + 1 + x + y^2}{(x + y + z)^2} = \frac{3 + x + y + z + x^2 + y^2 + z^2}{(x + y + z)^2}$$

5. Therefore, it suffices to prove that:
$$(x + y + z)^2 \geq 3 + x + y + z + x^2 + y^2 + z^2$$

6. Expanding and rearranging terms, we get:
$$x^2 + y^2 + z^2 + 2(xy + yz + zx) \geq 3 + x + y + z + x^2 + y^2 + z^2$$
$$2(xy + yz + zx) \geq 3 + x + y + z$$

7. Using the given condition $xy + yz + zx = x + y + z$, we substitute:
$$2(x + y + z) \geq 3 + x + y + z$$
$$x + y + z \geq 3$$

8. We know that $3(x + y + z) = 3(xy + yz + zx) \leq (x + y + z)^2$, which implies:
$$x + y + z \geq 3$$

9. Equality holds if $x = y = z = 1$, as this satisfies both the given condition and the inequality.

$\blacksquare$