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

Question

Let $a,b,c\ge 0$ such that $a^2+b^2+c^2=1$. Prove that $$\sqrt{1-\frac{\left(a+b\right)^2}{4}}+\sqrt{1-\frac{\left(b+c\right)^2}{4}}+\sqrt{1-\frac{\left(a+c\right)^2}{4}}\ge \sqrt{6}$$

---

My try:

$$\sqrt{1-\frac{\left(a+b\right)^2}{4}}=\sqrt{a^2+b^2+c^2-\frac{\left(a+b\right)^2}{4}}=\frac{\sqrt{3a^2-2ab+3b^2+4c^2}}{2}\ge\frac{\sqrt{2a^2+2b^2+4c^2}}{2}$$$

Then i need to prove $$\sum \sqrt{2a^2+2b^2+4c^2}\ge 2\sqrt{6}$$

But that's failed. Without using $a^2+b^2\ge 2ab$ to remove the radical i have no more idea, i tried to use holder but failed. Help me.

Answer 1

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

Answer 2

Also, Minkowski (triangle inequality) helps.

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

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