$$ \text{Waring's problem asks, "Is }\left\lfloor \left(\frac{3}{2}\right)^n\right\rfloor =\left\lfloor \frac{3^n-1}{2^n-1}\right\rfloor\text{ always true?"} $$
We craft an inequality, with $m,n \in\mathbb{Z}^{+}$,
$$ \frac{(2 m-1) 3^n-1}{(2 m-1) 2^n-1}<\left\lceil \left(\frac{3}{2}\right)^n\right\rceil, $$ where $(2m-1)$ is the $m$-th odd number. Note: when $m=1$, this is the floor item as shown at top right.
Some representative results: Edit to set domain to integers $$ m\in \mathbb{Z}\land n=1\land m\geq 2 \\ m\in \mathbb{Z}\land n=2\land m\geq 1 \\ m\in \mathbb{Z}\land n=3\land m\geq 1 \\ m\in \mathbb{Z}\land n=400\land m\geq 1 \\ $$ Take $n\rightarrow\infty$ and we have Waring's Problem solved.
Q: Do we need more?
Q2: Can you spot the error in my answer?