Proving Inequality for All $n \ge 1$

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Discussion Overview

The discussion centers around proving the inequality $$ \frac{n^3}{n^5 + 4n + 1} \le \frac{1}{n^2}$$ for all integers $n \ge 1$. Participants explore various approaches to establish the validity of this inequality, including algebraic manipulations and conditions under which the inequality holds.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions how the inequality can be guaranteed when $n^3$ is in the numerator, asking if this holds for any numerator greater than 1.
  • Another participant suggests rearranging the inequality and states that the condition $n^5 + 4n + 1 > n^5$ leads to $n > -\frac{1}{4}$, noting that for integers, this implies $n \ge 1$.
  • A different participant argues that if the inequality holds, it leads to the conclusion that $0 \le 4n + 1$, which is true for positive $n$.
  • One participant reiterates the need to prove the original statement and provides a step-by-step breakdown of the reasoning, confirming that the inequality holds for $n > 1$.

Areas of Agreement / Disagreement

Participants express varying degrees of confidence in the inequality's validity, with some providing supportive reasoning while others raise questions about the conditions under which it holds. There is no clear consensus on the necessity of the conditions or the implications of the inequality.

Contextual Notes

Participants discuss the implications of the inequality for different ranges of $n$, particularly focusing on the restriction to $n \ge 1$. There is also mention of the undefined nature of the right-hand side when $n = 0.

Who May Find This Useful

This discussion may be of interest to those studying inequalities in mathematics, particularly in the context of algebraic manipulation and proof strategies for inequalities involving polynomial expressions.

tmt1
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I have this inequality:

$$ \frac{n^3}{n^5 + 4n + 1} \le \frac{1}{n^2}$$

for all $n \ge 1$

I get that

$$ \frac{1}{n^5 + 4n + 1} \le \frac{1}{n^2}$$

but how do I guarantee that when $n^3$ is in the numerator, this inequality holds? Is this for any numerator greater than 1? Also, why must $n$ be greater than or equal to 1?
 
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Rearrange the left-hand side like so:

$\dfrac{n^2}{n^2}\cdot \dfrac{n^3}{n^5 + 4n + 1} = \dfrac{1}{n^2}\cdot\dfrac{n^5}{n^5 + 4n + 1} < \dfrac{1}{n^2}$

whenever $n^5 + 4n + 1 > n^5$, that is, when $4n + 1 > 0$, so $n > -\frac{1}{4}$.

If $n$ is an integer, this means $n$ must be non-negative. But we cannot allow $n = 0$, or else the RHS of the inequality is undefined. That leaves $n \geq 1$ (unless you want to make some awkward qualifications about when $n = 0$).

The inequality still holds for all non-zero reals $n$ greater than $-\frac{1}{4}$, but the use of the letter $n$ typically indicates a natural number.
 
Since n is positive, if if were true that $$\frac{n^3}{n^5+ 4n+ 1}\le \frac{1}{n^2}$$ then, multiplying by [math]n^2(n^5+ 4n+ 1)[/math] we would have [math]n^5\le n^5+ 4n+ 1[/math]. That is the same as [math]0\le 4n+ 1[/math] which, since n is positive, is true. To prove the original statement, work back. It is true that [math]0\le 4n+ 1[/math]. Add [math]n^5[/math] to both sides to get [math]n^5\le n^5+ 4n+ 1[/math]. Now divide both sides by [math]n^2(n^4+ 4n+ 1)[/math].
 
tmt said:
I have this inequality: $ \frac{n^3}{n^5 + 4n + 1} \le \frac{1}{n^2} $ for all $n \ge 1$
I get that.

$$ \frac{1}{n^5 + 4n + 1} \le \frac{1}{n^2}$$

but how do I guarantee that when $n^3$ is in the numerator, this inequality holds?
Is this for any numerator greater than 1?
Also, why must $n$ be greater than or equal to 1?
\begin{array}{cccc}\text{For } n &gt; 1, &amp; 4n + 1 \:\ge\:0 \\ \\<br /> \text{Add }n^5: &amp; n^5 + 4n+1 \:\ge\:n^5 \\ \\<br /> \text{Take reciprocals:} &amp; \dfrac{1}{n^5+4n+1} \:\le \:\dfrac{1}{n^5} \\ \\<br /> \text{Multiply by }n^3: &amp; \dfrac{n^3}{n^5+4n+1} \:\le \: \dfrac{n^3}{n^5} \\ \\<br /> \text{Therefore:} &amp; \dfrac{n^3}{n^5+4n+1} \:\le\: \dfrac{1}{n^2}<br /> \end{array}
 

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