how much will k=0 get, while using some (very basic) results of the triangle inequality

@above i doubt that'll get partial

WLOG let $k\geq 0.$ The main idea is to let $b_i=a_i-ka_{i+1}-a_{i+2}$ for $0\leq i \leq n-2$ and write \begin{align*}(a_0-ka_1-a_2)w_0&=b_0w_0\\ (a_1-ka_2-a_3)w_1&=b_1w_1 \\ &\vdots \\ (a_{n-2}-ka_{n-1}-a_n)w_{n-2}&=b_{n-2}w_{n-2}\end{align*}and sum these equations where $w_i-kw_{i+1}-w_{i+2}=0$ for each $i.$ By the theory of characteristic polynomials, a natural choice is $w_i=r^i$ where $r$ is a root of $x^2-kx-1=0,$ and then summing will result in $a_0-w_{n-2}a_n=\sum_{i=0}^{n-2}b_iw_i.$ If we pick $r$ to be the negative root, $r=\tfrac{k-\sqrt{k^2+4}}{2}\in [-1,0)$ and so $w_i\in[-1,1]$ for each $i.$ Now the rest is just an application of the triangle inequality: \begin{align*} |a_0|-|a_n|&\leq |a_0|-|w_{n-2}||a_n|\\ &\leq |a_0-w_{n-2}a_n|\\ &=\left|\sum_{i=0}^{n-2}b_iw_i\right|\\&\leq \sum_{i=0}^{n-2}|b_i||w_i|\leq \sum_{i=0}^{n-2}|b_i|, \end{align*}done. Unfortunately my solution involved choosing $w_i$ such that $w_0=w_n=1,$ which might still work but is a bit messier. Any guesses on partials for this?

um how do you like.... think of that. am i just bad at algebra or is this standard

Motivation: using the triangle inequality, we want to bound $\left|\sum b_i\right|$ and the easiest way to do that is to add weights to make things telescope (the given $a_1=a_{n-1}=0$ particularly suggests this, as these terms won't cancel with the rest).

Dose anybody have an answer？It's too hard for me

Generic_Username wrote: WLOG let $k\geq 0.$ The main idea is to let $b_i=a_i-ka_{i+1}-a_{i+2}$ for $0\leq i \leq n-2$ and write \begin{align*}(a_0-ka_1-a_2)w_0&=b_0w_0\\ (a_1-ka_2-a_3)w_1&=b_1w_1 \\ &\vdots \\ (a_{n-2}-ka_{n-1}-a_n)w_{n-2}&=b_{n-2}w_{n-2}\end{align*}and sum these equations where $w_i-kw_{i+1}-w_{i+2}=0$ for each $i.$ By the theory of characteristic polynomials, a natural choice is $w_i=r^i$ where $r$ is a root of $x^2-kx-1=0,$ and then summing will result in $a_0-w_{n-2}a_n=\sum_{i=0}^{n-2}b_iw_i.$ If we pick $r$ to be the negative root, $r=\tfrac{k-\sqrt{k^2+4}}{2}\in [-1,0)$ and so $w_i\in[-1,1]$ for each $i.$ Now the rest is just an application of the triangle inequality: \begin{align*} |a_0|-|a_n|&\leq |a_0|-|w_{n-2}||a_n|\\ &\leq |a_0-w_{n-2}a_n|\\ &=\left|\sum_{i=0}^{n-2}b_iw_i\right|\\&\leq \sum_{i=0}^{n-2}|b_i||w_i|\leq \sum_{i=0}^{n-2}|b_i|, \end{align*}done. Unfortunately my solution involved choosing $w_i$ such that $w_0=w_n=1,$ which might still work but is a bit messier. Any guesses on partials for this? Wonderful solution!

I'll present a solution similar to above, with a bit different way. Let $t$ be a root of $x^2 - kx - 1 = 0$ with $|t|\geq 1$. Then, the other root is $-\frac{1}{t}$, so $k = t - \frac{1}{t}$. So, for each $i = 0, 1, \ldots, n-2$, $$ |a_i - ka_{i+1} - a_{i+2}| = \left| (a_i - ta_{i+1}) + \frac{1}{t}(a_{i+1} - ta_{i+2}) \right| \geq |a_i - ra_{i+1}| - \frac{1}{|t|} |a_{i+1} - ta_{i+2}| \geq |a_i - ta_{i+1}| - |a_{i+1} - ta_{i+2}| $$ since $\frac{1}{|t|}\leq 1$. Summing up from $i = 0$ to $n - 2$ using the inequality above (preserving $\frac{1}{|t|}$ for $i = n - 2$), $$ \sum_{i = 0}^{n-2} |a_i - ka_{i+1} - a_{i+2}| \geq |a_0 - ta_1| - \frac{1}{|t|}|a_{n-1} - ta_n| = |a_0| - |a_n| $$

BlazingMuddy wrote: I'll present a solution similar to above, with a bit different way. Let $t$ be a root of $x^2 - kx - 1 = 0$ with $|t|\geq 1$. Then, the other root is $-\frac{1}{t}$, so $k = t - \frac{1}{t}$. So, for each $i = 0, 1, \ldots, n-2$, $$ |a_i - ka_{i+1} - a_{i+2}| = \left| (a_i - ta_{i+1}) + \frac{1}{t}(a_{i+1} - ta_{i+2}) \right| \geq |a_i - ra_{i+1}| - \frac{1}{|t|} |a_{i+1} - ta_{i+2}| \geq |a_i - ta_{i+1}| - |a_{i+1} - ta_{i+2}| $$ since $\frac{1}{|t|}\leq 1$. Summing up from $i = 0$ to $n - 2$ using the inequality above (preserving $\frac{1}{|t|}$ for $i = n - 2$), $$ \sum_{i = 0}^{n-2} |a_i - ka_{i+1} - a_{i+2}| \geq |a_0 - ta_1| - \frac{1}{|t|}|a_{n-1} - ta_n| = |a_0| - |a_n| $$ Nice solution.

BlazingMuddy wrote: I'll present a solution similar to above, with a bit different way. Let $t$ be a root of $x^2 - kx - 1 = 0$ with $|t|\geq 1$. Then, the other root is $-\frac{1}{t}$, so $k = t - \frac{1}{t}$. So, for each $i = 0, 1, \ldots, n-2$, $$ |a_i - ka_{i+1} - a_{i+2}| = \left| (a_i - ta_{i+1}) + \frac{1}{t}(a_{i+1} - ta_{i+2}) \right| \geq |a_i - ra_{i+1}| - \frac{1}{|t|} |a_{i+1} - ta_{i+2}| \geq |a_i - ta_{i+1}| - |a_{i+1} - ta_{i+2}| $$ since $\frac{1}{|t|}\leq 1$. Summing up from $i = 0$ to $n - 2$ using the inequality above (preserving $\frac{1}{|t|}$ for $i = n - 2$), $$ \sum_{i = 0}^{n-2} |a_i - ka_{i+1} - a_{i+2}| \geq |a_0 - ta_1| - \frac{1}{|t|}|a_{n-1} - ta_n| = |a_0| - |a_n| $$ I didn't get how you got the last line

Generic_Username wrote: Prove that for $n>1$ and real numbers $a_0,a_1,\dots, a_n,k$ with $a_1=a_{n-1}=0$, \[|a_0|-|a_n|\leq \sum_{i=0}^{n-2}|a_i-ka_{i+1}-a_{i+2}|.\] Write each\[a_i - ka_{i+1} - a_{i+2} = (a_i + ca_{i+1}) - \tfrac 1c (a_{i+1} + ca_{i+2})\]for some $c$ satisfying $\tfrac 1c - c = k \implies 1 - c^2 = ck \implies c = \tfrac{-k \pm \sqrt{k^2 + 4}}{2}$. Take the one with more magnitude, clearly $\geq 1$ for all real $k$, hence $\left|\tfrac 1c\right| \leq 1$. Then we may write\begin{align*}\sum_{i = 0}^{n-2} |(a_i + ca_{i+1}) \pm \left|\tfrac 1c\right|(a_{i+1} + ca_{i+2})| &\geq \sum_{i=0}^{n-2} |a_i + ca_{i+1}| - \left|\tfrac 1c\right||a_{i+1} + ca_{i+2}|\\&\geq \left(\sum_{i = 0}^{n-3} |a_i + ca_{i+1}| - |a_{i+1} + ca_{i+2}|\right) + (|a_{n-2} + ca_{n-1}| - \left|\tfrac 1c\right||a_{n-1} + ca_{n}|)\end{align*}as $\left|\tfrac 1c\right| \leq 1$. This nicely telescopes to $|a_0| - |a_n|$ as desired. $\blacksquare$