Proving Inequality for All $n \ge 1$

[tmt1]

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?

 

[Deveno]

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.

 

[HallsofIvy]

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].

 

[soroban]

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?

$$\begin{array}{cccc}\text{For } n > 1, & 4n + 1 \:\ge\:0 \\ \\<br /> \text{Add }n^5: & n^5 + 4n+1 \:\ge\:n^5 \\ \\<br /> \text{Take reciprocals:} & \dfrac{1}{n^5+4n+1} \:\le \:\dfrac{1}{n^5} \\ \\<br /> \text{Multiply by }n^3: & \dfrac{n^3}{n^5+4n+1} \:\le \: \dfrac{n^3}{n^5} \\ \\<br /> \text{Therefore:} & \dfrac{n^3}{n^5+4n+1} \:\le\: \dfrac{1}{n^2}<br /> \end{array}$$