how to prove $\sum_{cyc}(1-a_{1}+a_{1}a_{2})^2\ge\frac{n}{2}$

Question

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?

Answer 1

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 $$abcd+1\geq0.$$

Answer 2

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 $$f_{n}=(1-a_{1}+a_{1}a_{2})^2+(1-a_{2}+a_{2}a_{3})^2+\cdots+(1-a_{n}+a_{n}a_{1})^2\ge\dfrac{9n}{16}$$

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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.

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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 !

Answer 3

I would say that you can not prove it using induction.

Firstly, let us rewrite $f(n)$ as $$f_{n}=(1+a_{1}(a_{2}-1))^2+(1+a_{2}(a_{3}-1))^2+\cdots+(1+a_{n}(a_{1}-1))^2$$ By inductive hypothesis, $$(1+a_{1}(a_{2}-1))^2+(1+a_{2}(a_{3}-1))^2+\cdots+(1+a_{n}(a_{1}-1))^2\ge\dfrac{n}{2}$$

Therefore, we have that $$(1+a_{1}(a_{2}-1))^2+(1+a_{2}(a_{3}-1))^2+\cdots+(1+a_{n}(a_{n+1}-1))^2+(1+a_{n+1}(a_{1}-1))^2\geq$$

$$\dfrac{n}{2}-(1+a_{n}(a_{1}-1))^2+(1+a_{n}(a_{n+1}-1))^2+(1+a_{n+1}(a_{1}-1))^2$$

We have to prove that $$(1+a_{n}(a_{n+1}-1))^2+(1+a_{n+1}(a_{1}-1))^2-(1+a_{n}(a_{1}-1))^2\geq\frac{1}{2}$$

Unfortunately, this inequality is false (take for instance $a_n=0$, $a_{n+1}=1$, and $a_1$ whatever value you want lesser than $\frac{3}{4}$).