By AM-GM, $\sum \sqrt[3]{7 a^2 b +1 } = \frac{1}{4} \sum\sqrt[3]{8a\cdot8a\cdot( 7 b + \frac{1}{a^2}} )$ $\leq \frac{1}{12} \sum (8a+8a+7 b + \frac{1}{a^2})=2\sum a.$

From the AM-GM we get:$a+b+c\geq\frac{9}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ and from the QM-AM we get $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq \sqrt{3(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})}$.Then we have:$a+b+c\geq\frac{9}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\geq\frac{3\sqrt3}{\sqrt{\sum\frac{1}{a^2}}}$.From this with easy manipulations we get $a+b+c\geq 3$ or \[\sum \frac{1}{a^2}\geq 3\],for every a,b,c such that $a+b+c=\frac{1}{a^2}+...+\frac{1}{c^2}$. If we assume $abc> 1$ then there \[\sum\frac{1}{a^2}\geq 3\sqrt[3]{\frac{1}{a^2b^2c^2}}\] there are some numbers for which \[\sum \frac{1}{a^2}<3\] which is impossible and with the sam observation we an conclude $abc=1$. Then we have \[\sum_{cyc}\sqrt[3]{7a^2b+1}=\sum_{cyc} \sqrt[3]{7\cdot \frac{a}{c}+1}\leq \sum\sqrt[3]{8\cdot\frac{a}{c}}\].Where last statement is true by rearrangement. From that it follows:\[\sum\sqrt[3]{8\frac{a}{c}}=\sum 2\sqrt[3]{\frac{a}{c}\leq 2(a+b+c).\](We have to prove that). \[\sum 2\sqrt[3]{\frac{a}{c}\leq 2(a+b+c)\] \[\sqrt[3]{\frac{a}{c}}+\sqrt[3]{\frac{b}{a}+\sqrt[3]{c}{b}\leq a+b+c \] \[3\sqrt[3]{\frac{a}{c}}+3\sqrt[3]{\frac{b}{a}+\sqrt[3]{c}{b}\leq 3a+3b+3c=\sum a^2bc+\frac{1}{ab}+c^2 \] And that is true by AM-GM beacue $a^2bc+\frac{1}{ab}+c^2\geq 3\sqrt[3]{\frac{a}{c}}$.And we are done. $Q.E.D$

no,no satrujabog ! your solution is wrong. By am-gm $ a+b+c \ge 3(abc)^{\frac{1}{3}} $ and we have $ a+b+c \ge 3 $ ,but we don't know that $ abc \ge 1 $ or $ abc \le 1 $ .

Can you explain me something:If i got,using manipulations,for all a,b,c $a+b+c\geq 3$ and if its $abc<1$ then there are some numbers for which $a+b+c<3$ because:$a+b+c\geq 3(\sqrt[3]abc)$ and $3(abc)^{\frac{1}{3}}<3$ so there must be some a,b,c such that $a+b+c<3$ but isnt that impossible?

To find the case of equality, rather consider $\sum\sqrt[3]{7a^2b+1}=\frac14\sum\sqrt[3]{8a\cdot 8b\cdot \left(7a+\frac1{ab}\right)}\le$ $\le \frac{23}{12}(a+b+c)+\frac1{12}\sum\frac1{ab}\le 2(a+b+c)$, as the AM-GM implies we need $a=b=c$.

Apologies for reviving the thread, but a more "inspired" solution (for the inequality part) is as follows: $$8(a+b+c) = \sum_{cyc} \bigg(\frac{1}{a^2} + 7b \bigg) = \sum_{cyc} \frac{7a^2b+1}{a^2}$$Therefore, we have by Holder: $$8(a+b+c)^3 = \bigg( \sum_{cyc} \frac{7a^2b+1}{a^2} \bigg) \cdot (a+b+c)^2 \geq \bigg( \sum_{cyc}\sqrt[3]{7a^2b+1} \bigg)^3$$Taking cube roots gives the desired inequality.

We can use that $\sqrt[3]{8}=2$, so factorize it. Later, we will use $AM-GM$ inequality $\sqrt[3]{7a^2b+1}=2\sqrt[3]{a^2(\frac{7b}{8}+\frac{1}{8a^2})} \leq \frac{2}{3}(2a+\frac{7b}{8}\frac{1}{8a^2})$ $\sqrt[3]{7b^2c+1}=2\sqrt[3]{b^2(\frac{7c}{8}+\frac{1}{8b^2})} \leq \frac{2}{3}(2b+\frac{7c}{8}\frac{1}{8b^2})$ $\sqrt[3]{7c^2b+1}=2\sqrt[3]{c^2(\frac{7a}{8}+\frac{1}{8c^2})} \leq \frac{2}{3}(2c+\frac{7a}{8}\frac{1}{8c^2})$. Now, add up this to get: $\sqrt[3]{7a^2b+1}+\sqrt[3]{7b^2c+1}+\sqrt[3]{7c^2b+1} \leq \frac{2}{3}(2a+2b+2c+\frac{7a+7b+7c}{8}+\frac{1}{8}(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}))$. After using problem facts and simple factorization, we will get: $2(a+b+c) \geq \sqrt[3]{7a^2b +1}+\sqrt[3]{7b^2c +1} + \sqrt[3]{7c^2a +1}$ Equality is when $a=b=c=1$

we make the most of the condition of the matter.we look for $x$ and $y$ as follows. $${7a^2b+1}=(xa)(ya)(\frac {7b+ \frac{1}{a^2}}{xy})$$$$\sqrt[3]{7a^2b+1} \leq \frac{xa+ya+ \frac {7b+ \frac{1}{a^2}}{xy}}{3}$$$$ \sum_{cyc} \sqrt[3]{7a^2b+1} \leq \frac{(x+y)(a+b+c)+ \frac{7}{xy}(a+b+c) + \frac{1}{xy} (\frac {1}{a^2}+\frac {1}{b^2}+\frac {1}{c^2})}{3}= \frac {(x+y)(a+b+c)+ \frac{7}{xy}(a+b+c)+ \frac{1}{xy}(a+b+c)}{3} = (\frac{x+y+ \frac{8}{xy}}{3})(a+b+c)$$$\frac{x+y+ \frac{8}{xy}}{3}=2$ it suffices to specify an $x$ and $y$ that satisfy this condition.for example (2.2)

Different and easier solution $$ a+b+c=\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \iff (a+b+c)a^2b^2c^2=a^2b^2+b^2c^2+c^2a^2 \ge abc(a+b+c) \implies abc \ge 1$$So by Holder inequality $$\sum_{cyc}\sqrt[3]{7a^2b+1} \le \sum_{cyc} \sqrt[3]{7a^2b+abc}=\sum_{cyc} \sqrt[3]{ab(7a+c)} \le \sqrt[3]{(\sum_{cyc}a)^2 (\sum_{cyc}(7a+c)) }=2(a+b+c) \qquad \blacksquare$$