MathNet 2025 — Part 01 (100 complete bilingual questions) Question 15

For all $a$, $b$, $c$ positive real numbers with $a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 1$, show that
$$
\frac{ab + bc + ca}{1 + 2abc} \le \frac{(a + b + c)^2}{4}.
$$