My solution:The inequality is equalent to: $\prod\limits_{cyc} (1+\frac{ab+ac+bc}{2a\sqrt{bc} +bc}) \ge 8$ Now using the inequality $\sqrt{bc} \le \frac{b+c}{2}$ we get the result we needed. Second solution(Not mine):apply holders inequality then use the solution of H.Hafezi2000. Third solution(Not mine):Use mixing variables method.

Taha1381 wrote: My solution:The inequality is equalent to: $\prod\limits_{cyc} (1+\frac{ab+ac+bc}{2a\sqrt{bc} +bc}) \ge 8$ Now using the inequality $\sqrt{bc} \le \frac{b+c}{2}$ we get the result we needed. I don't think. It reduces to that

Taha1381 wrote: For positive real integers $ab+ac+bc=1$ prove that: $\prod\limits_{cyc} (\sqrt{bc}+\frac{1}{2a+\sqrt{bc}}) \ge 8abc$ $ (\sqrt{bc}+\frac{1}{2a+\sqrt{bc}}) \ge 2 \sqrt{bc} $ which is obvious and we're done! It's just the idea here: https://artofproblemsolving.com/community/u366128h1694327p10850620 at post 3 the first solution

Pluto1708 wrote: Taha1381 wrote: My solution:The inequality is equalent to: $\prod\limits_{cyc} (1+\frac{ab+ac+bc}{2a\sqrt{bc} +bc}) \ge 8$ Now using the inequality $\sqrt{bc} \le \frac{b+c}{2}$ we get the result we needed. I don't think. It reduces to that Solution is correct.

I know now

Note that : $$ab+ac \stackrel {AM-GM}{\geq} 2a\sqrt {bc}$$$$ \implies \sum_{cyc} bc \geq 2a\sqrt {bc}+ bc$$$$ \implies \frac {1}{2a+ \sqrt {bc}} \geq \sqrt {bc}$$ Hence we have : $$ \prod_{cyc} \left( \sqrt bc + \frac {1}{2a+ \sqrt {bc}} \right) \geq \prod_{cyc} \left( 2 \sqrt {bc} \right)=8abc$$ We are done. $\blacksquare$

If we get $ab=x^2 , ac=y^2 , bc=z^2$ , $\sum x^2 =1$ , it's easy to check that we should just say $8\Pi(2xy+z^2) \leq \Pi (2xy+z^2+1)$ if we get $2xy+z^2=s , 2yz+y^2=t , 2yz+x^2=r $ , $\sum s \leq 3 \longrightarrow srt \leq 1$ so we should say $8srt \leq \Pi(s+1) \leftrightarrow 7srt \leq \sum st +\sum s +1 $ we know $\sum s +\sum st +1 \geq 7 \sqrt[7]{s^3r^3t^3} \geq 7 srt $ and the problem is solved $\blacksquare$

We have \begin{align*} \prod_{cyc} \left(\sqrt{bc}+\frac{1}{2a+\sqrt {bc}}\right)=\prod_{cyc} \left(\frac{2a\sqrt {bc}+2bc+ab+ca}{2a+\sqrt {bc}}\right)\geqslant \prod_{cyc} \left(\frac{4a\sqrt {bc}+2bc}{2a+\sqrt {bc}}\right)=\prod_{cyc} 2\sqrt {bc}=8abc. \end{align*}

Generalization 1 Let $a_{1},a_{2},\cdots,a_{n}$ be positive reals ($n\geq 3$) such that $\sum_{cyc}{a_{1}a_{2}}=\lambda $. Then prove that $$\prod_{cyc-j}{\left(\sqrt{\dfrac{\prod{a_{1}}}{a_{j}}}+\dfrac{\lambda }{2.\sum\limits_{k=j}^{j+n-3}{\left(\dfrac{a_{k}}{\sqrt{\dfrac{\prod{a_{1}}}{a_{k-1}a_{k}a_{k+1}}}}\right)}+a_{j-2}a_{j-1}\sqrt{\dfrac{a_{j}}{\prod{a_{1}}}}}\right)}\geq 2^n\sqrt{\left(\prod{a_{1}}\right)^{n-1}}$$