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
 A: 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)
A: I) We have $2x+y=12-(y+z)-z \leq 12-5-3=4$. Thus by AM-GM inequality $2xy \leq (\frac{2x+y}{2})^2 \leq 4$. Finally by QM-AM inequality $$x^2+y^2+z^2=(x+y)^2+z^2-2xy \geq 2(\frac{(x+y)+z}{2})^2-2xy=18-2xy \geq 14$$
II) Let $s=\sqrt{x}+\sqrt{y}+\sqrt{z}$. Then $$s^2=x+y+z+2(\sqrt{xy}+\sqrt{xz}+\sqrt{yz})=6+2(\sqrt{xy}+\sqrt{xz}+\sqrt{yz})$$
$$(\frac{s^2-6}{2})^2=xy+xz+yz+2s\sqrt{xyz}$$
We have by I) $xy+xz+yz=\frac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=\frac{36-(x^2+y^2+z^2)}{2} \leq 11$. Also, $6x+3y+2z=36-3(y+z)-z \leq 36-3(5)-3=18$ so by AM-GM inequality $36xyz \leq (\frac{6x+3y+2z}{3})^3 \leq 216$, so $xyz \leq 6$. Thus 
$$(\frac{s^2-6}{2})^2 \leq 11+2\sqrt{6}s$$
$$s^4 \leq 12s^2+8+8\sqrt{6}s=12(s+\frac{\sqrt{6}}{3})^2$$
$$s^2 \leq 2\sqrt{3}(s+\frac{\sqrt{6}}{3})=2\sqrt{3}s+2\sqrt{2}$$
$$(s-\sqrt{3})^2 \leq 2\sqrt{2}+3=(1+\sqrt{2})^2$$
$$s \leq 1+\sqrt{2}+\sqrt{3}$$
