RO-NMO 2015 Problem 94

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