Question 53
Let $a$, $b$, $c$ be real numbers in the interval $[0, 1]$ with $a + b$, $b + c$, $c + a \ge 1$. Prove that
$$
1 \le (1-a)^2 + (1-b)^2 + (1-c)^2 + \frac{2\sqrt{2abc}}{\sqrt{a^2+b^2+c^2}}.
$$
$$
1 \le (1-a)^2 + (1-b)^2 + (1-c)^2 + \frac{2\sqrt{2abc}}{\sqrt{a^2+b^2+c^2}}.
$$