First, substitution: $$xy=a+b,yz=b+c,zx=c+a, \lbrace a,b,c\rbrace\subset \mathbb{R}_+$$We're left with \begin{align*} \sum \frac{\sqrt{a+b}}{\sqrt{2c+(k+1)(a+b)}} \geq 2 \sqrt{1-k} \end{align*}Now we take a look on: $$ \frac{\sqrt{a+b}}{\sqrt{2c+(k+1)(a+b)}}\ge \sqrt{1-k}\cdot\frac{a+b}{a+b+c}$$It's equivalent to $$\left(c+k(a+b)\right)^2\ge0$$so it's true. Needless to say: $$\sum \frac{\sqrt{a+b}}{\sqrt{2c+(k+1)(a+b)}}\ge \sum\sqrt{1-k}\cdot\frac{a+b}{a+b+c}=2\sqrt{1-k}$$with equality iff $ck-c=bk-b=ak-a=k(a+b+c)$ which is $a=b=c, k=-\frac12$ and finally $$x=y=z, k=-\frac12$$Remark If $k\ge 0$ you can get a better bounding $$ \frac{\sqrt{a+b}}{\sqrt{2c+(k+1)(a+b)}}\ge \frac{a+b}{(a+b+c)\sqrt{k+1}}$$equivalent to $$2k(a+b)c+(k+1)c^2\ge0$$and you'll have $$\sum\frac{\sqrt{a+b}}{\sqrt{2c+(k+1)(a+b)}}\ge\sum \frac{a+b}{(a+b+c)\sqrt{k+1}}=\frac{2}{\sqrt{k+1}}\ge2\sqrt{1-k}$$