$2a^2+\frac{1}{a}=a^2+a^2+\frac{1}{a}\geq 3a$. So we must prove that $$\sum{\frac{a}{b+\frac{1}{a}+1}}\geq 1.$$ By Cauchy-Schwarz $$\sum{\frac{a}{b+\frac{1}{a}+1}}=\sum{\frac{a^2}{ab+1+a}}\geq \frac{(a+b+c)^2}{ab+bc+ca+a+b+c+3}.$$ It remains to prove that $$a^2+b^2+c^2+ab+bc+ca\geq a+b+c+3.$$ $$ab+bc+ca\geq 3\sqrt[3]{a^2b^2c^2}=3 \ \star$$ $$a^2+b^2+c^2\geq \frac{(a+b+c)(a+b+c)}{3}\geq \frac{3(a+b+c)}{3}=a+b+c \ \ \star\star$$ With summing up $\star$ and $\star\star$ we get desired result. Note: This is also Azerbaijan JBMO TST 2014

IstekOlympiadTeam wrote: $2a^2+\frac{1}{a}=a^2+a^2+\frac{1}{a}\geq 3a$. So we must prove that $$\sum{\frac{a}{b+\frac{1}{a}+1}}\geq 1.$$ By Cauchy-Schwarz $$\sum{\frac{a}{b+\frac{1}{a}+1}}=\sum{\frac{a^2}{ab+1+a}}\geq \frac{(a+b+c)^2}{ab+bc+ca+a+b+c+3}.$$ It remains to prove that $$a^2+b^2+c^2+ab+bc+ca\geq a+b+c+3.$$ Which is easy by AM-GM I have another solution but it is Long your solution is the best

Murad.Aghazade wrote: $\boxed{\text{A7}}$ Let $a,b,c$ be positive reals such that $abc=1$.Prove the inequality $\sum\frac{2a^2+\frac{1}{a}}{b+\frac{1}{a}+1}\geq 3$ See also here http://www.artofproblemsolving.com/community/c6h1077228p4713586

Let $a = \frac xy$, $b = \frac yz$, $c = \frac zx$. The desired inequality becomes $$\sum_{\text{cyc}} \frac{\frac{2x^2}{y^2}+\frac yx}{\frac yz+\frac yx+\frac yy} \ge 3$$ which rearranges to $$\sum_{\text{cyc}} \left( \frac{2x^2}{y^3} + \frac 1x \right) \ge 3 \left( \frac1x + \frac1y + \frac1z \right)$$ or $$\sum_{\text{cyc}} \frac{x^2}{y^3} \ge \frac1x + \frac1y + \frac1z$$ which follows by AM-GM since $$4 \cdot \frac{x^2}{y^3} + 6 \cdot \frac{y^2}{z^3} + 9 \cdot \frac{z^2}{x^3} \ge 19 \sqrt[19]{\frac{x^2}{y^3} \cdot \frac{y^2}{z^3} \cdot \frac{z^2}{x^3}} = 19 \cdot \frac1x.$$

Setting $(a, b, c) = \left(\frac yz, \frac zx, \frac xy\right),$ the inequality simplifies to $$\sum \frac{xy^3}{z^2} \geq \sum xy,$$ which is clear say by Rearrangement on $\{1/z^2\}$ and $\{xy^3\}$.

IstekOlympiadTeam wrote: $2a^2+\frac{1}{a}=a^2+a^2+\frac{1}{a}\geq 3a$. So we must prove that $$\sum{\frac{a}{b+\frac{1}{a}+1}}\geq 1.$$ By Cauchy-Schwarz $$\sum{\frac{a}{b+\frac{1}{a}+1}}=\sum{\frac{a^2}{ab+1+a}}\geq \frac{(a+b+c)^2}{ab+bc+ca+a+b+c+3}.$$ It remains to prove that $$a^2+b^2+c^2+ab+bc+ca\geq a+b+c+3.$$ $$ab+bc+ca\geq 3\sqrt[3]{a^2b^2c^2}=3 \ \star$$ $$a^2+b^2+c^2\geq \frac{(a+b+c)(a+b+c)}{3}\geq \frac{3(a+b+c)}{3}=a+b+c \ \ \star\star$$ With summing up $\star$ and $\star\star$ we get desired result. Note: This is also Azerbaijan JBMO TST 2014 Nice, last step follows from WAM-WHM.

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Note that $$\sum_{\text{cyc}} \frac{2a^2 + \frac1a}{b + \frac1a + 1} = \sum_{\text{cyc}} \frac{2a^2}{bc + b + 1} + \sum_{\text{cyc}} \frac{1}{bc + b + 1}$$ We now consider the individual parts. Claim: The first part is at least $2$. Proof. By Titu's, $$\sum_{\text{cyc}} \frac{2a^2}{bc + b + 1} \ge \frac{2(a + b + c)^2}{ab + bc + ca + (a + b + c)(abc)^{\frac13} + 3(abc)^{\frac23}}$$ Substituting in $a + b + c = 1$ and $(abc)^{\frac{1}{3}} \le \frac{1}{3}$, we get that the sum is at least $$\frac{2}{\frac13 + \frac13 + \frac13} = 2.$$ $\blacksquare$ Claim: The second part is exactly $1$. Proof. This expands as $$\begin{align*} &\sum_{\text{cyc}} \frac{1}{ab + a + 1} = \frac{1}{ab + a + 1} + \frac{1}{\frac{1}{a} + b + 1} + \frac{1}{\frac{1}{b} + \frac{1}{ab} + 1} \\ &= \frac{1}{ab + a + 1} + \frac{a}{ab + a + 1} + \frac{ab}{ab + a + 1} = 1 \end{align*}$$ $\blacksquare$

We first note that $$a^2+b^2+c^2 \ge \frac{(a+b+c)^2}{3} \ge \frac{\sqrt[3]{abc} \cdot (a+b+c)}{3} = a+b+c,$$ $$2(ab+bc+ca) \ge 3(abc)^{2/3} + (ab+bc+ca) = 3abc + ab+bc+ca.$$ Summing these, we have $$(a+b+c)^2 \ge 3abc+(ab+bc+ca)+(a+b+c).$$ We finish by using Titu's and AM-GM on the numerator: $$1 \leq \frac{(a+b+c)^2}{(a+b+c)+(ab+bc+ca)+3abc} \leq \sum \frac{a}{b+bc+abc} \leq \sum \frac{\frac 13 (a^2+a^2+\frac 1a)}{b+\frac 1a + 1}.$$ $$\implies \sum \frac{2a^2+\frac 1a}{b+\frac 1a + 1} \ge 3.~\blacksquare$$

After substituting $a=y/x,b=z/y,c=x/z$ it remains to prove $\frac{y^3}{x^2}+\frac{z^3}{y^2}+\frac{x^3}{z^2}\ge x+y+z.$ By AM-GM we have $$\frac4{19}\left(\frac{y^3}{x^2}\right)+\frac6{19}\left(\frac{z^3}{y^2}\right)+\frac9{19}\left(\frac{x^3}{z^2}\right)\ge \left(\frac{y^3}{x^2}\right)^{\frac4{19}}\cdot\left(\frac{z^3}{y^2}\right)^{\frac6{19}}\cdot\left(\frac{x^3}{z^2}\right)^{\frac9{19}}=x$$ and summing cyclically finishes.