My textbook says that, for [imath]|x|>1[/imath] and if [imath]n\in\mathbb{N}[/imath] is big enough, it is [imath]|x|^n \ge 2[/imath] and this implies that [imath]\frac{1}{1+|x|^n} \le \frac{2}{|x|^n}[/imath]. It is simple to prove this inequality by just multiplying for [imath]|x|^n[/imath] both sides, however I would like to understand how the author deduces this inequality. I thought that I can prove this by introducing a generic parameter [imath]a[/imath] that I can fix later, and try to prove [imath]\frac{1}{1+|x|^n} \le \frac{a}{|x|^n}[/imath] using the hypothesis that [imath]|x|^n \ge 2[/imath]. If I assume [imath]a \ge 1[/imath], I can say [imath]\frac{1}{1+|x|^n} \le \frac{a}{|x|^n} \iff |x|^n \ge \frac{a}{a-1}[/imath]. Since I know that [imath]|x|^n \ge 2[/imath], I can solve [imath]2\ge \frac{a}{a-1}[/imath] and so [imath]a \ge 2[/imath]. Hence [imath]a=2[/imath] is the optimal value and I deduce the inequality of the textbook. Is this reasoning valid?
An inequality
- Thread starter Ozma
- Start date
blamocur
Elite Member
- Joined
- Oct 30, 2021
- Messages
- 3,222
But if [imath]|x|^n \geq 2[/imath] then [imath]\frac{1}{1-|x|^n}[/imath] is negative, isn't it?
@blamocur: Sorry if I come back on this, but I still have doubts about this. In the following, I have written this without mistakes. Can you gently check if this is correct, please?
My textbook says that if [imath]n \in \mathbb{N}[/imath] is big enough and [imath]|x|>1[/imath], then [imath]|x|^n\ge2[/imath] and from this it says (without further explanation) that [imath]\frac{1}{|x|^n-1}\le\frac{2}{|x|^n}[/imath]. I can of course prove the inequality backwards, however, since the author is interested in an estimation of the kind [imath]\frac{1}{|x|^n-1}\le\frac{k}{|x|^n}[/imath] for some [imath]k\in\mathbb{R}[/imath], I would like to understand the process that the author of textbook has used to find that inequality without knowing what constant works at the numerator.
So I tried this: for [imath]|x|^n\ge2[/imath], it is [imath]\frac{1}{|x|^n-1}\le\frac{k}{|x|^n} \iff k(|x|^n-1)\ge|x|^n \iff |x|^n(k-1) \ge k[/imath]. Since I am interested in finding some [imath]k\in\mathbb{R}[/imath] that work, I can impose that [imath]k>1[/imath] so that [imath]|x|^n(k-1) \ge k \iff |x|^n\ge\frac{k}{k-1}[/imath].
But, in this situation, it is [imath]|x|^n\ge2[/imath] and so, if I impose [imath]2\ge\frac{k}{k-1}[/imath] with [imath]k>1[/imath], then I will find all the [imath]k>1[/imath] that work. For [imath]k>1[/imath], it is [imath]2\ge\frac{k}{k-1} \iff k\ge2[/imath], and so if [imath]|x|^n \ge 2[/imath] it is [imath]\frac{1}{|x|^n-1}\le\frac{k}{|x|^n}[/imath] for any [imath]k\ge2[/imath]. Hence, since for any [imath]k\ge2[/imath] it is [imath]\frac{2}{|x|^n}\le\frac{k}{|x|^n}[/imath], the optimal value is for [imath]k=2[/imath] and I get the inequality from my textbook.
Is this reasoning a correct way to deduce this inequality and, in general, is introducing a generic parameter, solve for it (adding conditions if necessary, as I did in this situation) and finally fix the value of the parameters a correct technique to deduce inequalities?
My textbook says that if [imath]n \in \mathbb{N}[/imath] is big enough and [imath]|x|>1[/imath], then [imath]|x|^n\ge2[/imath] and from this it says (without further explanation) that [imath]\frac{1}{|x|^n-1}\le\frac{2}{|x|^n}[/imath]. I can of course prove the inequality backwards, however, since the author is interested in an estimation of the kind [imath]\frac{1}{|x|^n-1}\le\frac{k}{|x|^n}[/imath] for some [imath]k\in\mathbb{R}[/imath], I would like to understand the process that the author of textbook has used to find that inequality without knowing what constant works at the numerator.
So I tried this: for [imath]|x|^n\ge2[/imath], it is [imath]\frac{1}{|x|^n-1}\le\frac{k}{|x|^n} \iff k(|x|^n-1)\ge|x|^n \iff |x|^n(k-1) \ge k[/imath]. Since I am interested in finding some [imath]k\in\mathbb{R}[/imath] that work, I can impose that [imath]k>1[/imath] so that [imath]|x|^n(k-1) \ge k \iff |x|^n\ge\frac{k}{k-1}[/imath].
But, in this situation, it is [imath]|x|^n\ge2[/imath] and so, if I impose [imath]2\ge\frac{k}{k-1}[/imath] with [imath]k>1[/imath], then I will find all the [imath]k>1[/imath] that work. For [imath]k>1[/imath], it is [imath]2\ge\frac{k}{k-1} \iff k\ge2[/imath], and so if [imath]|x|^n \ge 2[/imath] it is [imath]\frac{1}{|x|^n-1}\le\frac{k}{|x|^n}[/imath] for any [imath]k\ge2[/imath]. Hence, since for any [imath]k\ge2[/imath] it is [imath]\frac{2}{|x|^n}\le\frac{k}{|x|^n}[/imath], the optimal value is for [imath]k=2[/imath] and I get the inequality from my textbook.
Is this reasoning a correct way to deduce this inequality and, in general, is introducing a generic parameter, solve for it (adding conditions if necessary, as I did in this situation) and finally fix the value of the parameters a correct technique to deduce inequalities?
blamocur
Elite Member
- Joined
- Oct 30, 2021
- Messages
- 3,222
This is a very good point! People often ask: what made you figure this out? Tell me what to do so I can make a good guess too. My only advice is start with simple problems before moving on to more difficult ones.
Cubist
Senior Member
- Joined
- Oct 29, 2019
- Messages
- 1,702
I too would use a substitution. The following seems very easy to follow IMO...
Substitute u=|x|^n, and we are given u ≥ 2
We need an inequality in the form of 1/(u-1) ≤ k/u, rearrange this to find a suitable value for k
k/u ≥ 1/(u-1)
k ≥ u/(u-1)
Substitute v=u-1 (and u≥2 implies v≥1)
k ≥ (v+1)/v
k ≥ 1 + 1/v
k ≥ 1 + 1 (this is the worst case scenario since v≥1)
k ≥ 2
Choose k=2 giving 1/(u-1) ≤ 2/u