Let $k\ge 1$. I want to prove that for all $a,b,c\geq 0$ we have $$\sqrt{k^2 a^2+bc}+\sqrt{k^2 b ^2+ac}+\sqrt{k^2 c ^2+ab}\le\left(k+\frac12\right)\cdot(a+b+c).$$
My attempt: I tried to use that $\sqrt{k^2 a^2+bc}\le ka+\sqrt{bc}$ and got (by AM-GM) $$\sum_{cyc} \sqrt{k^2 a^2+bc}\le\sum_{cyc}ka+\sqrt{bc}\le\sum_{\text{cyc}}ka+\frac{b+c}2=(k+1)\cdot(a+b+c).$$
However, this leaves me with an extra $\frac{a+b+c}2$ on the right hand side and I don't see how I can improve my bound.