Prove that $\sum _{cyc}\sqrt{1-\frac{\left(a+b\right)^2}{4}}\ge \sqrt{6}$

By C-S $$\sum_{cyc}\sqrt{1-\frac{(a+b)^2}{4}}=\frac{1}{2}\sum_{cyc}\sqrt{3a^2-2ab+3b^2+4c^2}=$$ $$=\frac{1}{2}\sqrt{\sum_{cyc}(10a^2-2ab+2\sqrt{(3a^2-2ab+3b^2+4c^2)(3a^2-2ac+3c^2+4b^2)}}=$$ $$=\frac{1}{2}\sqrt{\sum_{cyc}\left(10a^2-2ab+2\sqrt{((a-b)^2+2a^2+2b^2+2c^2+2c^2)((a-c)^2+2a^2+2b^2+2c^2+2b^2)}\right)}\geq$$ $$\geq\frac{1}{2}\sqrt{\sum_{cyc}(10a^2-2ab+2((a-b)(a-c)+2a^2+2b^2+2c^2+2bc))}=\frac{1}{2}\sqrt{\sum_{cyc}24a^2}=\sqrt6.$$

Also, Minkowski (triangle inequality) helps.

Indeed, let $a\geq b\geq c$.

Thus, $$\sum_{cyc}\sqrt{1-\frac{(a+b)^2}{4}}=\frac{1}{2}\sum_{cyc}\sqrt{3a^2-2ab+3b^2+4c^2}=$$ $$=\frac{1}{2}\sum_{cyc}\sqrt{(a-b)^2+2+2c^2}\geq\frac{1}{2}\sqrt{(a-b+a-c+b-c)^2+2\cdot3^2+2(a+b+c)^2}=$$ $$=\sqrt{(a-c)^2+5+ab+ac+bc}=\sqrt{6+(a-b)(b-c)}\geq\sqrt6.$$