Inequality Parameter

First, notice that plugging in $x=y=z=1$ gives $3+k\ge 6$ , so $k\ge 3$ . Now, we will prove that $3$ is indeed the minimum.

By AM-GM, $\dfrac{x+y+z+3}{2}=\dfrac{(x+2)+(y+z+1)}{2}\ge\sqrt{(x+2)(y+z+1)}$ By Cauchy-Schwarz, $(x+2)(y+z+1)=(x+1+1)(1+y+z)\ge (\sqrt{x}+\sqrt{y}+\sqrt{z})^2$ and plugging into the inequality from AM-GM, $\dfrac{x+y+z+3}{2}\ge\sqrt{x}+\sqrt{y}+\sqrt{z}$ which simplifies to the desired inequality.

Motivation: Having multiple square roots on the lesser side is troublesome, a single one could be obtained by AM-GM, but we also realize that Cauchy-Schwarz, in effect, takes square roots. Then, we must obtain another square root, which is possible through AM-GM.

Let $x= a^2$ , $y= b^2$ , $z= c^2$ . The inequality in question can be re-arranged as: $(a^2-2a) + (b^2-2b) + (c^2-2c) \geq -k$ Observe that $a^2-2a= (a-1)^2 - 1$ This shows that the minimum possible value of $a^2-2a$ is $-1$ , which is attained for $a= 1$ . Similarly, $b^2-2b \geq -1$ $c^2-2c \geq -1$ Adding them, $(a^2- 2a)+(b^2-2b)+(c^2-2c) \geq -3$ Equality is attained for $(a, b, c)= (1, 1, 1)$ . We thus conclude our answer is $\boxed{3}$ .

Let $f(x,y,z) = 2\sqrt{x} -x + 2\sqrt{y}-y+2\sqrt{z}-z$ . If we take $grad f = 0$ , then we obtain:

$f_{x} = \frac{1}{\sqrt{x}} - 1 =0, f_{y} = \frac{1}{\sqrt{y}} -1 =0, f_{z} = \frac{1}{\sqrt{z}} - 1 = 0 \Rightarrow x=y=z=1$ (i).

The Hessian Matrix at this critical point evaluates to:

$F(x,y,z) = \begin{bmatrix} f_{xx} & f_{xy} & f_{xz} \\ f_{yx} & f_{yy} & f_{yz} \\ f_{zx} & f_{zy} & f_{zz} \end{bmatrix} = \begin{bmatrix} -\frac{1}{2x^{3/2}} & 0 & 0 \\ 0 & -\frac{1}{2y^{3/2}} & 0 \\ 0 & 0 & -\frac{1}{2z^{3/2}}\end{bmatrix}$ ;

or $F(1,1,1) = -\frac{1}{2} \cdot I_{3x3}$ , which is negative-definite (hence, $P(1,1,1)$ is the global maximum point of $f$ over $x,y,z \in \mathbb{R^{+}}$ ). Thus, $k \ge f(1,1,1) = 3(2\sqrt{1} -1) \Rightarrow \boxed{k \ge 3}.$