Prove $x^2+y^2+z^2 \ge 14$ with constraints Let $0<x\le y \le z,\ z\ge 3,\ y+z \ge 5,\ x+y+z = 6.$ Prove the inequalities:
$I)\ x^2 + y^2 + z^2 \ge 14$
$II)\ \sqrt x + \sqrt y + \sqrt z \le 1 + \sqrt 2 + \sqrt 3$
My teacher said the method that can solve problem I can be use to solve problem II. But I don't know what method that my teacher talking about, so the hint is useless, please help me. 
Thanks

Hint:
$$x^2+y^2+z^2 \ge 14 = 1^2+2^2+3^2\iff (x-1)(x+1)+(y-2)(y+2)+(z-3)(z+3) \ge 0$$
$$\iff (z-3)[(z+3)-(y+2)] + (y+z-5)[(y+2)-(x+1)] + (a+b+c-6)(a+1) \ge 0$$ (alway true)
