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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

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    $\begingroup$ is there any additional condition given? $\endgroup$ – Dr. Sonnhard Graubner Jun 26 '17 at 16:44
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    $\begingroup$ from where does it come? $\endgroup$ – Dr. Sonnhard Graubner Jun 26 '17 at 16:47
  • $\begingroup$ The inequality is homogeneous hence you may assume $a+b+c=1$ without loss of generality. If $f(x)=\sqrt{\frac{2}{x}-3}$ were convex on $\left(0,\frac{2}{3}\right)$ the conclusion would be straightforward from Jensen's inequality. It is not the case, but with the substitution $a=\frac{2}{3}-\frac{2}{3}\tilde{a}^2$ such approach works. $\endgroup$ – Jack D'Aurizio Jun 26 '17 at 16:59
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By AM-GM $$\sum_{cyc}\sqrt{\frac{2b+2c}{a}-1}=\sum_{cyc}\frac{2\sqrt3(2b+2c-a)}{2\sqrt{3a(2b+2c-a)}}\geq$$ $$\geq\sum_{cyc}\frac{2\sqrt3(2b+2c-a)}{3a+2b+2c-a}=\sum_{cyc}\frac{2\sqrt3(2b+2c-a)}{2(a+b+c)}=3\sqrt3$$

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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!

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  • $\begingroup$ @inequality What do you think about my solution? Thank you! $\endgroup$ – Michael Rozenberg Jun 27 '17 at 1:57

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