Question 18
Let $a, b, c, d \ge 0$ be real numbers such that $a + b + c + d = 1$. Prove that
$$
\sqrt{a + \frac{(b-c)^2}{6} + \frac{(c-d)^2}{6} + \frac{(d-b)^2}{6}} + \sqrt{b} + \sqrt{c} + \sqrt{d} \le 2.
$$
$$
\sqrt{a + \frac{(b-c)^2}{6} + \frac{(c-d)^2}{6} + \frac{(d-b)^2}{6}} + \sqrt{b} + \sqrt{c} + \sqrt{d} \le 2.
$$