Question 54
Prove that for any positive real numbers $a$, $b$, $c \lt 1$ satisfying
$$
2(a+b+c) + 4abc = 3(ab+bc+ca) + 1,
$$
the following inequality holds: $a+b+c \le \frac{3}{4}$.
$$
2(a+b+c) + 4abc = 3(ab+bc+ca) + 1,
$$
the following inequality holds: $a+b+c \le \frac{3}{4}$.