Inequality Revisited

By AM-GM Inequality , we have

$x+y \geq 2\sqrt{xy}$

$\Rightarrow x+y+z \geq 2\sqrt{xy}+z$

$\Rightarrow 1 \geq 2\sqrt{xy}+z$

$\Rightarrow z \geq 2z\sqrt{xy}+z^2$

$\Rightarrow xy+z\geq xy + z^2 + 2z\sqrt{xy}$

$\Rightarrow xy+z \geq \left(\sqrt{xy} + z\right)^2$

$\Rightarrow \sqrt{xy+z} \geq \sqrt{xy} + z$

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Similarly ,

$\sqrt{yz+x} \geq \sqrt{yz} +x$

$\sqrt{zx+y} \geq \sqrt{zx} + y$

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Adding the three inequalities gives ,

$\sqrt{xy+z} + \sqrt{yz+x}+\sqrt{zx+y} \geq 1+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}$

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Therefore $\boxed{k=1}$ with equality when $\boxed{x=y=z=\dfrac13}$