Absolute Inequality

Algebra Level 4

Consider the sequence with real terms $a_{0},a_{1},a_{2},...,a_{k}$ that satisfies the inequality $\sum_{i=1}^{k} |a_{i-1}+a_{i}|≤\frac{3}{4}-3|a_{k}|$ . The maximum possible value of $|a_{0}a_{k}|$ can be expressed as $\frac{m}{n}$ with co-prime positive integers $m,n$ . Find $m+n$ .

The answer is 137.

1 solution

Sam Zhou
Jun 5, 2019

As $|x+y|≥|x|-|y|$ for $x,y \in \mathbb{R}$ ,

$\frac{3}{4}≥\sum_{i=1}^{k} |a_{i-1}+a_{i}|+3|a_{k}|≥|a_{0}|-|a_{1}|+|a_{1}|-|a_{2}|+...-|a_{k}|+3|a_{k}|=|a_{0}|+2|a_{k}|$ .

Using the AM-GM inequality,

$\frac{3}{4}≥|a_{0}|+2|a_{k}|≥2\sqrt{|2a_{0}a_{k}|}$ , so $\frac{3}{8}≥\sqrt{|2a_{0}a_{k}|}$ .

Hence $|a_{0}a_{k}|≤\frac{9}{128}$ , giving the answer of $\boxed{137}$ .

You technically didn't show that 9/128 is possible

Razzi Masroor - 1 year, 5 months ago

Absolute Inequality

The answer is 137.

1 solution

0 pending reports