Cool square root problem

Note that, since $a+b+c = 1$ , the AM\GM inequality tells us that $\sqrt{a + bc} \; = \; \sqrt{(1-b)(1-c)} \; \le \; \tfrac12((1-b)+(1-c)) \; = \; 1 - \tfrac12(b+c)$ with equality when $b=c$ . Arguing similarly, $\sqrt{a + bc} + \sqrt{b+ac} + \sqrt{c + ab} \; \le \; 3 - \tfrac12(b+c) - \tfrac12(a+c) - \tfrac12(a+b) \; = \; 3 - (a+b+c) \; = \; 2$ with equality when $a=b=c=\tfrac13$ .