how to prove $\sum_{cyc}(1-a_{1}+a_{1}a_{2})^2\ge\frac{n}{2}$ let $a_{i}\in [0,1]$,prove or disprove
$$f_{n}=(1-a_{1}+a_{1}a_{2})^2+(1-a_{2}+a_{2}a_{3})^2+\cdots+(1-a_{n}+a_{n}a_{1})^2\ge\dfrac{n}{2}$$
I can only prove $n=3$.
$$f_{3}=\sum_{cyc}[1-a_{1}(1-a_{2})]^2=3-2\sum_{cyc}a_{1}(1-a_{2})+\sum_{cyc}a^2_{1}(1-a_{2})^2$$
let $$t=a_{1}(1-a_{2})+a_{2}(1-a_{3})+a_{3}(1-a_{1})=1-(1-a_{1})(1-a_{2})(1-a_{3})-a_{1}a_{2}a_{3}\le 1$$
and
$$\sum_{cyc}a^2_{1}(1-a_{2})^2=(\sum_{cyc}a_{1}(1-a_{2}))^2-2\sum_{cyc}a_{1}a_{2}(1-a_{1})(1-a_{3})\ge t^2-\dfrac{1}{4}\sum_{cyc}a_{2}(1-a_{3})=t^2-\dfrac{1}{4}t$$
so we have
$$f_{3}\ge 3-2t+t^2-\dfrac{1}{4}t\ge\dfrac{3}{2}$$
But for $n$ I can't,maybe use induction to prove it?

A proof for $n=4$.
Let $a_1=\frac{1}{a+1},$ $a_2=\frac{1}{b+1},$ $a_3=\frac{1}{c+1}$ and $a_4=\frac{1}{d+1},$ where $a\geq0$, $b\geq0,$ $c\geq0$ and $d\geq0.$
Thus, $$\prod_{cyc}(a+1)^2\left(\sum_{cyc}(1-a_1+a_1a_2)^2-2\right)=$$
$$=\prod_{cyc}(a+1)^2\left(\sum_{cyc}\left(\frac{a}{a+1}+\frac{1}{(a+1)(b+1)}\right)^2-2\right)=2+2a^2b^2c^2d^2+$$
$$+\sum_{cyc}(2a^2b^2c^2d+a^2b^2c^2+2a^2b^2cd+2a^2b^2c+a^2b^2+2a^2bc-2a^2cd+2a^2b+a^2+2ab+2a)\geq$$
$$\geq\sum_{cyc}(a^2b^2c^2-2a^2cd+a^2)=\sum_{cyc}a^2(cd-1)^2\geq0.$$
A proof for $n=5$ we can get by the same way, but it's more complicated.
I found a nice proof for $n=4$.
By using the previous substitutions and by AM-GM we obtain:
$$\sum_{cyc}(1-a_1+a_1a_2)^2=\sum_{cyc}\left(\frac{ab+a+1}{(a+1)(b+1)}\right)^2\geq$$
$$\geq2\left(\frac{(ab+a+1)(cd+c+1)}{(a+1)(b+1)(c+1)(d+1)}+\frac{(bc+b+1)(da+d+1)}{(b+1)(c+1)(d+1)(a+1)}\right)$$ and it's enough to prove that:
$$(ab+a+1)(cd+c+1)+(bc+b+1)(da+d+1)\geq(a+1)(b+1)(c+1)(d+1),$$ which is $$abcd+1\geq0.$$
