BMO Short List

The inequality is equivalent to $$12(a+b+c-1)\ge \sum _{\text{cyc}}4\sqrt{2\left(a+\frac{b}{c}\right)}.$$ From AM-GM inequality we have $$\sum _{\text{cyc}}2\sqrt{2\left(a+\frac{b}{c}\right)}\le \sum _{\text{cyc}}\left(2+a+\frac{b}{c}\right)=6+a+b+c+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}.$$ Again from AM-GM inequality we have $$\frac{2a}{b}+\frac{2b}{c}+\frac{2c}{a}\ge 3\sqrt[3]{\frac{2a}{b}\cdot \frac{2b}{c}\cdot \frac{2c}{a}}=3\sqrt[3]{8}=6.$$ Hence, $$\sum _{\text{cyc}}2\sqrt{2\left(a+\frac{b}{c}\right)}\le a+b+c+\frac{3a}{b}+\frac{3b}{c}+\frac{3c}{a}.(1)$$ Again, from AM-GM inequality we have $$\begin{array}{rl}\sum _{\text{cyc}}2\sqrt{2\left(a+\frac{b}{c}\right)}& =\sum _{\text{cyc}}2\sqrt{\frac{2a}{c}\left(c+\frac{b}{a}\right)}\le \sum _{\text{cyc}}\left(\frac{2a}{c}+c+\frac{b}{a}\right)(2)\\ & =a+b+c+\frac{3a}{c}+\frac{3b}{a}+\frac{3c}{b}.\end{array}$$ Adding (1) and (2) we get, $$\sum _{\text{cyc}}4\sqrt{2\left(a+\frac{b}{c}\right)}\le 2(a+b+c)+3\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right).$$ Now, using the assumption we have $$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-3=3(a+b+c-1).$$ Hence, $$\sum _{\text{cyc}}4\sqrt{2\left(a+\frac{b}{c}\right)}\le 11(a+b+c)-9.$$ So, it is enough to prove $$11(a+b+c)-9\le 12(a+b+c-1)⟺a+b+c\ge 3.$$ Last inequality is true since using $AM-HM$ and condition we have, $$\frac{a+b+c}{3}\ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=1\Rightarrow a+b+c\ge 3.$$ It is known that equality in $AM-HM$ is achieved only when $a=b=c$ and, since $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$, $a=b=c=1$. Clearly, for $a=b=c=1$ equality holds.