We have $\left(a_1a_2^2+1\right)^2\le\left(a_2^3+1\right)\left(a_2a_1^2+1\right),$ and so on. Multiply these and find that \[\prod_{cyc} \left(a_1^3+1\right)\cdot\prod_{cyc} \left(a_1a_2^2+1\right)\geq\left[\prod_{cyc}\left(a_2a_1^2+1\right)\right]^2.\]Similarly, we get \[\prod_{cyc} \left(a_1^3+1\right)\cdot\prod_{cyc}\left(a_2a_1^2+1\right)\ge\left[\prod_{cyc}\left(a_2a_1^2+1\right)\right]^2.\]And from now on everything is simple.

A liitle more help would be fine! thanks

So, we have established that $\prod_{cyc} \left(a_1^3+1\right)\cdot\prod_{cyc} \left(a_1a_2^2+1\right)\geq\left[\prod_{cyc}\left(a_2a_1^2+1\right)\right]^2,$ and $\prod_{cyc} \left(a_1^3+1\right)\cdot\prod_{cyc}\left(a_2a_1^2+1\right)\ge\left[\prod_{cyc}\left(a_2a_1^2+1\right)\right]^2.$ If $\prod_{i=1}^n\left(a_ia_{i+1}^2+1\right)\geq\prod_{i=1}^n \left(a_{i+1}a_i^2+1\right),$ then we are done with the second relation. Otherwise, the first relation saves us.

Prove that for any positive numbers $a_1,\dots,a_n$ $(n \geq 2)$ \[(a_1^3+1)(a_2^3+1)\cdots(a_n^3+1) \geq (a_1^2 a_2+1)(a_2^2 a_3+1)\cdots (a_n^2 a_1+1)\]

Just posting the problem as it appeared in the contest, for the resources section.

Cauchy - Schwartz inequality

Wow It's very easy by KARAMATA INEQUALITY! Let $a_i = e^{x_i}$. So, we have $(3x_1,\cdots,3x_n)\succ (2x_1+x_2,\cdots,2x_n+x_1),$ and also $f(x)=\ln (e^x+1)$ is convex. Hence\[\ln (e^{3x_1}+1)+\cdots+\ln (e^{3x_n}+1) \ge \ln (e^{2x_1+x_2}+1)+\cdots+\ln (e^{2x_n+x_1}+1),\]and \[ (a_1^3+1)\cdots(a_n^3+1) \ge (a_1^2 a_2+1)\cdots(a_n^2 a_1+1).\]

But Holder is easier: $\prod_{i=1}^n(a_i^3+1)^3=\prod_{i=1}^n\Bigl((a_i^3+1)^2(a_{i+1}^3+1)\Bigr)\geq\prod_{i=1}^n(a_i^2a_{i+1}+1)^3$.

Hello Sorry to ask but Can you tell me what is holder inequality here I ain't gettin how you are going about this. By what formula or derivation you are gettin can you explain me the case of holder one pl's .

Holder inequality here it's the following. For $a_i>0$, $b_i>0$, $\alpha>0$ and $\beta>0$ we have: $$(a_1+a_2)^{\alpha}(b_1+b_2)^{\beta}\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}.$$In our case $\alpha=2$ and $\beta=1$.