Very nice inequality. Lemma: For $x,a,b\in [0,1]$, such that $x\ge a,b$, following inequality holds: \[ (1-bx+b^2)(1-ax+x^2) \ge 1-ab+b^2 \] Proof: It is equivalent to: $(x-a)(x-b)(1-bx)\ge 0$. what is obviously true because of condition. $\square$ Now, back to our original inequality. For $n=2$, inequality is equivalent to lemma when $a=b$. Suppose that original inequality holds for some $n\in \mathbb{N}$ and we want to prove it for $n+1$. WLOG we can suppose that $x_{n+1}=\max\{x_1,...,x_{n+1}\}$ (we can do that because inequality is cyclic). Now, using lemma for $x_n=b, \; x_{n+1}=x$, and $x_1=a$, we have: $(1-x_nx_{n+1}+x_n^2)(1-x_{n+1}x_1+x_{n+1}^2)\ge (1-x_nx_1+x_n^2)$ Therefore: ${{\prod_{i=1}^{n+1}(1-x_ix_{i+1}+x_i^2})\ge \prod_{i=1}^{n}(1-x_ix_{i+1}+x_i^2}) \ge 1$ and step of induction is proved. Hence, inequality is proved. $\blacksquare$

and very nice proof! excellent

Let's take $ln$ of both parts since all the factors of our product are positive. It suffices to prove that $\sum_{i=1}^{n} ln(1-x_i \cdot x_{i+1}+x_{i+1}^2) \geq 0$. Claim: $ln(1-x_i \cdot x_{i+1}+x_{i+1}^2) \geq \frac{x_{i+1}^2}{2}- \frac{x_i^2}{2}$ (note that we will solve our problem via claim). Let $f(x)=ln(1-a \cdot x+x^2)-\frac{x^2}{2}+\frac{a^2}{2}, a,x \in [0,1]$. Then $f'(x)=\frac{2x-a}{1-ax+x^2}-x=\frac{2x-a-x+ax^2-x^3}{1-ax+x^2}=\frac{(1-x^2)(x-a)}{1-ax+x^2}$. So $f(x)$ is descending on $[0,a]$ and ascending on $[a, 1]$ and is zero at $a$. $\square$