can anyone solve this?

Are you sure it's true? Notice that $c_{n+1}=a_n$ $c_i=a_ir+a_{i-1}$ for $1 \le i \le n$ $c_0=a_0r$ So then what if you had $a_n=a_{n-1}=...=a_1=1$, $a_0=1000000000$ or some other really big number, and then $r=\frac{1}{n+2}$? Clearly, $a=1000000000$, and if $n+3<1000000000$, $c=c_0$, and then we have the inequality untrue...

K81o7, you are wrong. $ c=c_{1} $ and the ratio is smaller than $ 1 $.

like K81o7 said we have: $ c_0 =ra_0 $ $ c_i =ra_i+a_{i-1} $ for $ 1\le i\le n $ $ c_{n+1}= a_n $. $ \frac{c}{|c_0|+|c_1|+....+|c_n|}\leq 1/n+1 $. we will prove that $ a $ is not greater than $ |c_0|+|c_1|+....+|c_n| $ using induction. suppose that for $ n-1 $ is true. if $ a=|a_n|$ then in the last sum we put $ c_{n+1} $ in place of $ c_0 $. if $ a=|a_i| $ and $ i $ is different than $ n $ then $ |c_0|+|c_1|+....+|c_{n-1}| $ is not smaller than $ a $ and $ |c_n| $ is not smaller than $ 0 $ so it is true because for $ n-1 $ is true. now we only have to prove it for $ n=0 $. $ f(x)=b $ and $ g(x)=bx+br $ so $ a=|b| $ and $ c=|b|x $ where $ x $ is the maximum of $ 1 $ and $ |r| $.we have to prove $ x $ is not greater than 1 which is true. q.e.d.

anonymouslonely wrote: we have: if $ a=|a_i| $ and $ i $ is different than $ n $ then $ |c_0|+|c_1|+....+|c_{n-1}| $ is not smaller than $ a $ and $ |c_n| $ is not smaller than $ 0 $ so it is true because for $ n-1 $ is true. When you add $ |c_n| $ and $ |a_n| $ but at the same time $ |c_{n-1}| $ changes,thus you,cannot use your induction suppose.That is my qusetion.Have I got anything wrong?Forgive my poor English.

niyinchen wrote: Forgive my poor English. You don't have to write that on post

niyinchen wrote: anonymouslonely wrote: we have: if $ a=|a_i| $ and $ i $ is different than $ n $ then $ |c_0|+|c_1|+....+|c_{n-1}| $ is not smaller than $ a $ and $ |c_n| $ is not smaller than $ 0 $ so it is true because for $ n-1 $ is true. When you add $ |c_n| $ and $ |a_n| $ but at the same time $ |c_{n-1}| $ changes,thus you,cannot use your induction suppose.That is my qusetion.Have I got anything wrong?Forgive my poor English. I also have your question.But I find a new way to prove it.

Why- wrote: niyinchen wrote: anonymouslonely wrote: we have: if $ a=|a_i| $ and $ i $ is different than $ n $ then $ |c_0|+|c_1|+....+|c_{n-1}| $ is not smaller than $ a $ and $ |c_n| $ is not smaller than $ 0 $ so it is true because for $ n-1 $ is true. When you add $ |c_n| $ and $ |a_n| $ but at the same time $ |c_{n-1}| $ changes,thus you,cannot use your induction suppose.That is my qusetion.Have I got anything wrong?Forgive my poor English. I also have your question.But I find a new way to prove it. How can you prove it?