Proving an equality involving a product $\frac{(3n^2-2)(3n^2-1)(1-3n+3n^3)}{2}\ge1$

Question

How can I prove that:

$$\frac{(3n^2-2)(3n^2-1)(1-3n+3n^3)}{2}\ge1$$

When $n\in\mathbb{N}$ such that $n\ge2$.

I think that we need to prove that all terms are monotonically increasing functions. But how do I prove that?

Answer 1

Well, plug $n=2$ in and see that the inequality holds. Then observe that every factor is monotonically increasing for $n \ge 2$, note that $3n^3-3n=3n(n^2-1)$.

Answer 2

Each factor is ultimately increasing. Find their minima:

• For $3 n^2 - 2$ it's minimum is at $n = 0$, it is positive for $n > \sqrt{2} / \sqrt{3}$, for $n \ge 2$ it is at least $10$.
• For $3 n^2 - 1$ the the minimum is also at $n = 0$, for $n \ge 2$ it is at least $11$.
• For $3 n^2 - 3 n + 1$ the minimum is at $n = 1/2$, for $n \ge 2$ it is at least $7$.

Pulling all together, for $n \ge 2$ your expression is at least: $\begin{align*} \frac{10 \cdot 11 \cdot 7}{2} &= 385 \end{align*}$

Answer 3

If you just plug $n+2$ in the expression and simply develop each term you get:

$$f(n+2)=\frac 12(3n^2+12n+11)(3n^2+12n+10)(19+33n+18n^2+3n^3)$$

There are $+$ signs everywhere, so for $n\ge 0$ we have $f(n+2)>0$

Note: With the same method, you can have the same result for $f(n+1)$.

Answer 4

Even $$3n^{3}-3n+1=3n(n^{2}-1)+1\geq 9n+1$$ for $$n\geq 2$$ thus Note that $$\frac{9n+1}{2}\geq 4.5\geq 1$$. Note that for $$n\geq 2$$ $$3n^{2}-2\geq 1$$ and $$3n^{2}-1\geq 1$$. Thus $$\frac{(3n^{3}-3n+1)(3n^{2}-1)(3n^{2}-2)}{2}\geq \frac{9n+1}{2}*1*1\geq 4.5\geq 1$$

Answer 5

Note that

$$n\geq 2 \Rightarrow n^2-1 \geq 3$$

Hence,

\begin{eqnarray*} \frac{(3n^2-2)(3n^2-1)(1-3n+3n^3)}{2} & \geq & \frac{(3n^2-3)(3n^2-3)(3n^3-3n)}{2} \\ & = & \frac{3^3(n^2-1)^3 n}{2} \\ & \stackrel{n^2-1\geq 3}{\geq} & 3^6 \\ & \geq & 1 \end{eqnarray*}