Problem 1350

Solution:
By $AM-GM$ we have $2x^{2}+\frac{1}{x}=x^{2}+x^{2}+\frac{1}{x} \geq 3 \sqrt[3]{\frac{x^{4}}{x}}=3x$ for all $x>0$, so we have:

$\sum_{\text{cyc}} \frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1} \geq \sum_{cyc} \frac{3a}{1+b+bc}=3\left(\sum_{cyc} \frac{a^{2}}{1+a+ab}\right) \geq \frac{3(a+b+c)^{2}}{3+a+b+c+ab+bc+ca}$.

By $AM-GM$ we have $ab+bc+ca \geq 3$ and $a+b+c \geq 3$. But $3\left(a^{2}+b^{2}+c^{2}\right) \geq (a+b+c)^{2} \geq 3(a+b+c)$. So $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca \geq 3+a+b+c+ab+bc+ca$. Hence $\sum_{cyc} \frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1} \geq \frac{3(a+b+c)^{2}}{3+a+b+c+ab+bc+ca} \geq \frac{3(a+b+c)^{2}}{(a+b+c)^{2}}=3$.

Denote $a=\frac{y}{x}$, $b=\frac{z}{y}$ and $c=\frac{x}{z}$. We have $\frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1}=\frac{\frac{2 y^{2}}{x^{2}}+\frac{x}{y}}{\frac{z}{y}+\frac{x}{y}+1}=\frac{2 y^{3}+x^{3}}{x^{2}(x+y+z)}$.

Hence $\sum_{cyc} \frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1}=\frac{1}{x+y+z} \cdot \sum_{cyc} \frac{2 y^{3}+x^{3}}{x^{2}}=\frac{1}{x+y+z} \cdot\left(x+y+z+2 \sum_{cyc} \frac{y^{3}}{x^{2}}\right)$.

By Rearrangement Inequality we get $\sum_{\text{cyc}} \frac{y^{3}}{x^{2}} \geq x+y+z$.

So $\sum_{\text{cyc}} \frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1} \geq \frac{1}{x+y+z} \cdot(3x+3y+3z)=3$.