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?

Partial answer
Well I work on $a_i\in[0.5,1]$ and $n\geq 5$
Let $s:I^2\to R$ where $I\subseteq R$ is an interval , have the following properties : $\sigma(x)=s(x,x) $ is increasing on $I$ ,$t(x)=s(a,x)+s(x,b)$ is increasing on $[\max{(a,b)},\infty[\cap I$ .If $a_i$ is an n-tuple with elements in $I$ and if $a=\min(a_1,a_2,\cdots,a_n)$ then :
$$\sum_{i=1}^{n}s(a_i,a_{i+1})\geq n\cdot s(a,a)$$
This case is a direct application of the theorem above wich can be found in the Dictionary of Inequalities Second edition (2015) by Professor Peter Bullen p.65
In fact we get a stronger statement we have :
Let $a_{i}\in [0.5,1]$ then
$$f_{n}=(1-a_{1}+a_{1}a_{2})^2+(1-a_{2}+a_{2}a_{3})^2+\cdots+(1-a_{n}+a_{n}a_{1})^2\ge\dfrac{9n}{16}$$


In the case $n=3$ with the Op's constraint we have :
$$\left(0.5-\frac{1}{3}\right)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right)+(1-(a+b)/2+ab)^{2}+(1-(b+c)/2+cb)^{2}+(1-(c+a)/2+ac)^{2}\geq1.5$$
And :
$$\left(0.5-\frac{1}{3}\right)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right)+(1-(a+b)/2+ab)^{2}+(1-(b+c)/2+cb)^{2}+(1-(c+a)/2+ac)^{2}\leq (1-a+ab)^{2}+(1-b+cb)^{2}+(1-c+ac)^{2}$$
Again I think a generalisation is possible .By generalisation we have to find the coefficient wich allows the equality case in the refinement .To finish I would say we have to play with $(a_i,a_j)$ or in the case $n=3$ $(a,b),(a,c),(b,c)$ wich give us $ab+bc+ca$ so we have to find the good permutation in each order.


Last attempt :
For $a_i\in[0,0.5]$ we have :
$$\sqrt{n}\sqrt{(1-a_{1}+a_{1}a_{2})^2+(1-a_{2}+a_{2}a_{3})^2+\cdots+(1-a_{n}+a_{n}a_{1})^2}-n\sqrt{0.5}\geq n+\sum_{i=1}^{n}a_i(a_{i+1}-1)-n\sqrt{0.5}\geq n+n(-0.25)-n\sqrt{0.5}>0$$
Hope it helps !
