Here is a bashy solution: Set $\sum a=p, \sum ab=q, abc=r$ The inequality is equivalent to $2pq^2 +9 \geqslant 7q^2$ Knowing $p^2 \geqslant 3q$ it is enough to prove $2\sqrt{3} q^{\frac{5}{2}} +9 -7q^2 \geqslant 0$ we also know that $q^3 \geqslant 27r=27$ so $q \geqslant 3$ Setting $f(q)= 2\sqrt{3} q^{\frac{5}{2}} +9 -7q^2$, we look at $f’(q)=5\sqrt{3} q^{\frac{3}{2}} -14q$, which for $q \geqslant 3$ is positive so the minimum of $f(q)$ for $q \geqslant 3$ occurs at $q=3$. Setting q=3, equality is achieved. this concludes the proof

I can't believe I just wasted my $500^{\textbf{th}}$ post , anyways here's my solution . Rewrite the $\textbf{LHS}$ and : \[\underbrace{\frac{a+b+c}{3}+\cdots+\frac{a+b+c}{3}}_{6}+\frac{9}{(ab+bc+ca)^2}\ge 7\cdot \sqrt[7]{\frac{(a+b+c)^6}{3^4\cdot (ab+bc+ca)^2}}\ge 7\cdot \sqrt[7]{\frac{ab+bc+ca}{3}}\ge{7}\]$\textbf{Q.E.D.}$

Kamran011 wrote: I can't believe I just wasted my $500^{\textbf{th}}$ post , anyways here's my solution . Rewrite the $\textbf{LHS}$ and : \[\underbrace{\frac{a+b+c}{3}+\cdots+\frac{a+b+c}{3}}_{6}+\frac{9}{a+b+c}\ge 7\cdot \sqrt[7]{\frac{(a+b+c)^6}{3^4\cdot (ab+bc+ca)^2}}\ge 7\cdot \sqrt[7]{\frac{ab+bc+ca}{3}}\ge{7}\]. $\textbf{Q.E.D}$ is $\frac{9}{(ab+bc+ca)^2}$ not $\frac{9}{a+b+c}$

sharifymatholympiad wrote: Positive real numbers $a$,$b$,$c$ are given such that $abc=1$.Prove that $$2(a+b+c)+\frac{9}{(ab+bc+ca)^2}\geq7.$$

Attachments:

$2(a+b+c)+ \frac{9}{(ab+bc+ca)^2} \geq 2(a+b+c)+ \frac{81}{(a+b+c)^4} \geq 7$(?) $\frac{2(a+b+c)}{3} + 4\frac{a+b+c}{3}+\frac{81}{(a+b+c)^4} \geq 7 $(?) $a+b+c \geq 3$ $\blacksquare$