There was one question in a question set that I was attempting that went something like this:
The number of solutions of the following inequality
$$2^\frac{1}{\sin^2 x_2} \cdot 3^\frac{1}{\sin^2 x_3} \cdot\;\cdots\;\cdot n^\frac{1}{\sin^2 x_n} \leq n!$$ where $$x_i \; \in \; \bigl(0,4\pi \bigl)$$ for $$i \equiv 1,2,3 \ldots n$$
My approach: $$0\leq\sin^2 x\leq 1 $$
$$ 1\leq \frac{1}{\sin^2x} \leq \infty$$
Now,
$$n^k \geq n \; \forall \; k \; \in \; [1,\infty]$$
Thus we can see that for the inequality $$2^\frac{1}{\sin^2 x_2} \cdot 3^\frac{1}{\sin^2 x_3} \ldots n^\frac{1}{\sin^2 x_n} \leq n!$$
to hold, the only value of $\frac{1}{\sin^2x}$ that satisfies it, is $1$. That gives us $2$ values for $\sin^2x $ satisfying the inequality in the interval $[0,2\pi]$ and $4$ values in the interval $[0,4\pi]$,
Thus we can see for $n$ terms there are $4^n$ values, however $1^\frac{1}{\sin^2 s_1}$ is not in the inequality so we get, $4^{n-1}$ values satisfying the inequality.
Please tell me if my reasoning is correct and I feel like we can also use Logarithms here maybe, but I'm not sure, so I would like to know whether that would also work.