$\sqrt{ab+c}=\sqrt{ab+1-a-b}=\sqrt{(1-a)(1-b)}\leq \frac{2-a-b}{2}$$\sum_{}^{} \frac{a+b}{\sqrt{ab+c}}=\sum_{}^{}\frac{a}{\sqrt{ab+c}}+\sum_{}^{}\frac{b}{\sqrt{ab+c}}\geq \sum_{}^{}\frac{2a^2}{2a-a^2-ab}+\sum_{}^{}\frac{2b^2}{2b-ab-b^2}\geq \frac{4(a+b+c)^2}{2(a+b+c)-(a^2+b^2+c^2)-(ab+bc+ca)} =\frac{4(a+b+c)^2}{2(a+b+c)-(a+b+c)^2+ab+bc+ca}\geq \frac{4(a+b+c)^2}{2(a+b+c)-(a+b+c)^2+\frac{(a+b+c)^2}{3}}=3$
$\sqrt{ab+c}=\sqrt{ab+1-a-b}=\sqrt{(1-a)(1-b)}\leq \frac{2-a-b}{2}$$\sum_{}^{} \frac{a+b}{\sqrt{ab+c}}=\sum_{}^{}\frac{a}{\sqrt{ab+c}}+\sum_{}^{}\frac{b}{\sqrt{ab+c}}\geq \sum_{}^{}\frac{2a^2}{2a-a^2-ab}+\sum_{}^{}\frac{2b^2}{2b-ab-b^2}\geq \frac{4(a+b+c)^2}{2(a+b+c)-(a^2+b^2+c^2)-(ab+bc+ca)} =\frac{4(a+b+c)^2}{2(a+b+c)-(a+b+c)^2+ab+bc+ca}\geq \frac{4(a+b+c)^2}{2(a+b+c)-(a+b+c)^2+\frac{(a+b+c)^2}{3}}=\frac{4}{3}$
$\sqrt{ab+c}=\sqrt{ab+1-a-b}=\sqrt{(1-a)(1-b)}\leq \frac{2-a-b}{2}$$\sum_{}^{} \frac{a+b}{\sqrt{ab+c}}=\sum_{}^{}\frac{a}{\sqrt{ab+c}}+\sum_{}^{}\frac{b}{\sqrt{ab+c}}\geq \sum_{}^{}\frac{2a^2}{2a-a^2-ab}+\sum_{}^{}\frac{2b^2}{2b-ab-b^2}\geq \frac{4(a+b+c)^2}{2(a+b+c)-(a^2+b^2+c^2)-(ab+bc+ca)} =\frac{4(a+b+c)^2}{2(a+b+c)-(a+b+c)^2+ab+bc+ca}\geq \frac{4(a+b+c)^2}{2(a+b+c)-(a+b+c)^2+\frac{(a+b+c)^2}{3}}=3$