Let $a,b,c>0$ and $k \geq\ 0$ prove $\sum \limits_{cyc}\sqrt{\dfrac{a}{b+kc}}\geq min\left\{\dfrac{3}{\sqrt{k+1}},\dfrac{2}{\sqrt[4]{k}}\right\}$

Question

Here $\dfrac{3}{\sqrt{k+1}}\leq\dfrac{2}{\sqrt[4]{k}}$ if $k\geq\dfrac{49+9\sqrt{17}}{32}$

Now i got the proof when $\dfrac{3}{\sqrt{k+1}}$ is minimum here. But i can not prove this when $\dfrac {2}{\sqrt[4]{k}}$ is minimum.

By holder inequality i got this$$\left(\sum \limits_{cyc}\sqrt{\dfrac{a}{b+kc}} \right)^2\left(\sum \limits_{cyc}\ (b+kc)\right)\geq (a+b+c)^3\implies \left(\sum \limits_{cyc}\sqrt{\dfrac{a}{b+kc}} \right)^2 \geq \dfrac{(a+b+c)^3}{\left(\sum \limits_{cyc}\ (b+kc)\right)}=\dfrac{(a+b+c)^2}{k+1}$$Hence if $a+b+c\geq 3$ then $$\sum \limits_{cyc}\sqrt{\dfrac{a}{b+kc}}\geq \dfrac{3}{\sqrt{k+1}}$$ But the other case i can't prove.