Question about establishing an inequality

Question

I am having trouble figuring out an inequality that is supposed to be simple.

Let N be a natural number and $t\ge N$, $j$ is also a natural number.

Assume that $\frac{2\log t}{\log N} \le j \le \frac{3\log t}{\log N}$ and $n\ge N$. Then from this information how can we show that $n^{j-\frac{\log t}{\log N}} \ge N^{2/j}$?

From the assumption we have $j-\frac{\log t}{\log N} \ge \frac{\log t}{\log N}$ so we can get $n^{j-\frac{\log t}{\log N}} \ge N^{\frac{\log t}{\log N}}$. But $j/2 \ge \log t / \log N$. So how can we get the desired inequality?

Answer 1

Let $\alpha:=\frac{\log t}{\log N}\ge1.$ Wlog, $n=N,$ so we just need to check that $j-\alpha\ge\frac2j,$ i.e. $j^2-\alpha j-2\ge0,$ i.e. $j\ge\frac{\alpha+\sqrt{\alpha^2+8}}2.$ This stems from $2j-\alpha\ge3\alpha=\sqrt{\alpha^2+8\alpha^2}\ge\sqrt{\alpha^2+8}.$

Answer 2

Let's use the assumption $\displaystyle\frac{2\log t}{\log N} \le j \le \frac{3\log t}{\log N}$ $(1)$. Knowing that $t\ge N, N\in \mathbb{N}$, $N\ne 1$. Since $t\ge N>1$, that means $\displaystyle0<1<\frac{2\log t}{\log N}$. Taking the reciprocals in the inequality $(1)$ we get: $$\frac{\log N}{2\log t} \ge \frac{1}{j} \ge \frac{\log N}{3\log t}\implies\frac{\log N}{\log t} \ge \frac{2}{j} \ge \frac{2\log N}{3\log t} (2)$$

You already found that $\displaystyle n^{j-\frac{\log t}{\log N}} \ge N^{\frac{\log t}{\log N}}$. We know that $N>1$ and $\displaystyle t\ge N\implies \frac{\log t}{\log N}\ge\frac{\log N}{\log t}\ge \frac{2}{j}$, thus we can conclude $\displaystyle n^{j-\frac{\log t}{\log N}} \ge N^{\frac{\log t}{\log N}}\ge N^{\frac{\log N}{\log t}}\ge N^{\frac{2}{j}}$.