How prove this inequality $\sum_{cyc}\sqrt{\frac{2b+2c}{a}-1}\ge 3\sqrt{3}$

Question

Let $a,b,c>0 ,2b+2c-a\ge 0,2c+2a-b\ge 0,2a+2b-c\ge 0$ show that $$\sqrt{\dfrac{2b+2c}{a}-1}+\sqrt{\dfrac{2c+2a}{b}-1}+\sqrt{\dfrac{2a+2b}{c}-1}\ge 3\sqrt{3}$$

I try use AM-GM and Cauchy-Schwarz inequality and from here I don't see what to do

Answer 1

By AM-GM $$\sum_{cyc}\sqrt{\frac{2b+2c}{a}-1}=\sum_{cyc}\frac{2\sqrt3(2b+2c-a)}{2\sqrt{3a(2b+2c-a)}}\geq$$ $$\geq\sum_{cyc}\frac{2\sqrt3(2b+2c-a)}{3a+2b+2c-a}=\sum_{cyc}\frac{2\sqrt3(2b+2c-a)}{2(a+b+c)}=3\sqrt3$$

Answer 2

Since our inequality is homogeneous, we can assume that $a+b+c=3$.

Hence, $2b+2c-a=2(3-a)-a=3(2-a)\geq0$, which gives $\{a,b,c\}\subset(0,2]$.

Thus, we need to prove that $$\sum_{cyc}\sqrt{\frac{2(b+c)}{a}-1}\geq3\sqrt3$$ or $$\sum_{cyc}\sqrt{\frac{2(3-a)}{a}-1}\geq3\sqrt3$$ or $$\sum_{cyc}\sqrt{\frac{2}{a}-1}\geq3$$ or $$\sum_{cyc}\left(\sqrt{\frac{2}{a}-1}-1\right)\geq0$$ or $$\sum_{cyc}\frac{1-a}{\sqrt{a}(\sqrt{2-a}+\sqrt{a})}\geq0$$ or $$\sum_{cyc}\left(\frac{1-a}{\sqrt{a}(\sqrt{2-a}+\sqrt{a})}+\frac{a-1}{2}\right)\geq0$$ or $$\sum_{cyc}\frac{(a-1)(\sqrt{a(2-a)}+a-2)}{\sqrt{a}(\sqrt{2-a}+\sqrt{a})}\geq0$$ or

$$\sum_{cyc}\frac{(a-1)\sqrt{2-a}(\sqrt{a}-1)}{\sqrt{a}(\sqrt{2-a}+\sqrt{a})}\geq0$$ or $$\sum_{cyc}\frac{(a-1)^2\sqrt{2-a}}{\sqrt{a}(\sqrt{2-a}+\sqrt{a})(\sqrt{a}+1)}\geq0.$$ Done!