About the smallest value of B

Question

I am trying to solve this problem:
We know that there's a inequality: $$(3n-1)(n+B)\geq A(4n-1)n$$ When $A=\frac{3}{4}$, what is the smallest possible value of B.
So, what I did is that:
$$B\geq \frac{\frac{3}{4}n(4n-1)-n(3n-1)}{3n-1}$$ We can deduce that:
$$B\geq \frac{3n(4n-1)-4n(3n-1)}{4(3n-1)}$$ Expand and simplifies: $$B\geq \frac{1}{12-\frac{1}{n}}$$ We know that when n is greater, the denominator of the RHS would be greater, meaning that RHS would be smaller. So, the smallest value of RHS would result in the smallest value of LHS.
When $n \to \infty$: The RHS $\to$ $\frac{1}{12}$.
I thought that the smallest value of B should be $\frac{1}{12}$. But, it turns out to be $\frac{1}{8}$.
Note: $n\geq 1$, and n is integers.
May I know why my method doesn't work? thank you so much.

Answer 1

Because it should be $$B\geq\frac{1}{12-\frac{4}{n}}.$$

The minimal value of $\frac{1}{12-\frac{4}{n}}$ does not exist bu the infimum is equal to $\frac{1}{12}.$

Also, fo $n\geq1$ we obtain: $$\frac{1}{12-\frac{4}{n}}\leq\frac{1}{8}.$$

Thus, the minimal value of $B$ for which the inequality $$B\geq\frac{1}{12-\frac{4}{n}}$$ is true for any natural $n$ it's $$B=\frac{1}{8}.$$

Answer 2

Find the smallest and greatest value of $B\ \forall\ n\in\mathbb N$ such that $$(3n-1)(n+B)\geq \frac34(4n-1)n$$

$$B\geq\frac{1}{12-\frac{4}{n}}$$ Draw the graph $y$ vs $n$.

• $y=0$ for $n=0$.
• Asymptote $\Rightarrow\text{ greatest }y=\frac1{12}$ as $n\to +\infty$.
• Asymptote $y\to-\infty$ as $n\to \frac13$.
• Smallest $y=\frac1{8}$ for $n=1$.