The trick is in figuring out how to handle $|ab + a + b| \leq 1/2$.. I still haven't figured it out yet. However, one can write $\sum |a_i| \leq \Bigg| \sum a_i \Bigg| \leq \frac{1}{2} + Z$, and this looks like it could reasonably do it, if good enough bounds can be found for the symmetric polynomials. (where Z is the modulus of the 2nd symmetric polynomial of the $a_i$, plus the 3rd, 4th, .. to the nth.)

Singular wrote: The trick is in figuring out how to handle $|ab + a + b| \leq 1/2$.. I still haven't figured it out yet. However, one can write $\sum |a_i| \leq \Bigg| \sum a_i \Bigg| \leq \frac{1}{2} + Z$, and this looks like it could reasonably do it, if good enough bounds can be found for the symmetric polynomials. (where Z is the modulus of the 2nd symmetric polynomial of the $a_i$, plus the 3rd, 4th, .. to the nth.) did you reverse the sign of triangle inequality?

yes i did, oops.

oh,is this so hard?

Did anyone figure out how to do it ?

Yes, maybe someone has the official solution ?

yes,I have .and I will post it soon

sorry.I can't find the office solution... and I 'm very sorry for breaking one's promise

It doesnot mind but there are so many Chinese in the forum, noone has the solution?

Fortunately,I find the offical solution now.And I think this problem is hard.Some problem is easy if you know the solution,but I think this one not competely is. now let me post the offical solution. let $a_j+1=r_j e^{i \theta_j},|\theta_j| \leq \pi$ for $j=1,2,\cdots,n$ with the condition we have $|\prod_{j \in I} r_j e^{i \sum_{j \in I} \theta_j} -1 |\leq \frac 1 2$ $(*)$ we will prove a lemma first. $lemma$,if $r,\theta$ are two real number,$r>0,|\theta| \leq \pi$ and $|r e^{i \theta}-1|<\frac 1 2$ then we have : (1)$\frac 1 2 \leq r \leq \frac 3 2$ (2)$|\theta| \leq \frac{\pi}6$ (3)$|re^{i \theta}-1| \leq |r-1|+|\theta|$ proof of lemma (1),(2) can be proved in a graph easily(also you can prove it by algebra) (3) is because $|re^{i \theta}-1=|r \cos \theta +r\sin \theta-1|=|(r-1)(\cos \theta+i \sin \theta)+(\cos \theta-1+i \sin \theta)| \leq |r-1|+\sqrt{(\cos \theta-1)^2+\sin^2 \theta}=|r-1|+2|\sin \frac{\theta}{2}| \leq |r-1|+|\theta|$ so the lemma is proved. return to the orginal problem: with the lemma we have: $\frac 1 2 \leq |\prod_{j \in I} r_j | \leq \frac 3 2$ ,$|\sum_{j \in I}| \leq \frac{\pi}6$ and $|a_j|=|r_j e^{i \theta_j}-1| \leq |r_j-1|+|\theta_j|$, so $\sum^n_{j =1}|a_j| \leq \sum^n_{j =1}|r_j-1|+\sum^n_{j=1}|\theta_j| =\sum_{r_j \geq 1}|r_j-1|+\sum_{r_j < 1}|r_j-1|+\sum_{\theta_j \geq 0}|\theta_j|+\sum_{\theta_j < 0}|\theta_j|$ but we have $\sum_{r_j \geq 1}|r_j-1| =\sum_{r_j \geq 1}(r_j-1) \leq \prod_{r_j \geq 1}(1+r_j-1)-1(**) \leq \frac 3 2 -1=\frac 1 2$ $\sum_{r_j <1 }|r_j-1| =\sum_{r_j \geq 1}(1-r_j) \leq \prod_{r_j < 1}(1-(1-r_j))^(-1)-1 \leq(***) \frac 3 2 -1 \leq 2-1=1$ $\sum^n_{j=1}|\theta_j-1| =\sum_{\theta_j \geq 0}|\theta_j| -\sum_{\theta_j < 0}|\theta_j| \leq \frac{\pi} 6 + \frac{\pi} 6 = \frac{\pi} 3$ so $\sum^n_{j =1}|a_j| \leq \frac 1 2 +1+\frac{\pi} 3<3$ $(**)$ and $(***)$ can be proved by induction easily. so the problem are proved. maybe this is my longest post in mathlink. and I hope it is clear for you.

zhaobin wrote: $\sum_{r_j <1 }|r_j-1| =\sum_{r_j \geq 1}(1-r_j) \leq \prod_{r_j < 1}(1-(1-r_j))^(-1)-1 \leq(***) \frac 3 2 -1 \leq 2-1=1$ Zhaobin, I don't how to prove this inequality. (i'm refering to this one : $\sum \ldots \leq \prod_{r_j<1} \ldots$) It doesn't seem obvious by induction. Can you help me?

ok,with pleasure. but an another nice proof is coming now just notice $\frac 1 {1-x} \ge 1+x$ for $0<x<1$ so $\prod_{r_j < 1}(1-(1-r_j))^{-1}-1 \ge \prod_{r_j < 1}(1+(1-r_j))^{-1}-1 \ge \sum_{r_j \geq 1}(1-r_j)$ is it clear? and I think induction also works.

thanks. now it is clear. my misunderstanding came from that $-1$. i didn't realize it was the exponent