Floor function and inequality I'm trying to work through a problem and I need to find a nice way to show that
$$\lfloor x^2\rfloor +\lfloor2x\rfloor\leq\lfloor x^4\rfloor$$ 
for $x\in[\sqrt[4]{3},\infty)$.
I know one possible way is to break down that interval into all the places where the three floor values increase by $1$. However, I was wondering if there is a way to do this in 1-2 lines.
 A: Hint: by definition of the greatest integer function:
$$
\begin{cases}
\begin{align}
\lfloor x^2\rfloor +\lfloor2x\rfloor & \leq x^2 + 2 x \\
\lfloor x^4\rfloor & \gt x^4 - 1
\end{align}
\end{cases}
$$
Therefore a sufficient condition for the inequality to hold is:
$$
x^2+2x \le x^4 - 1 \;\; \iff \;\; x^4-x^2-2x-1 = x^4 - (x+1)^2 = (x^2-x-1)(x^2+x+1) \ge 0
$$
The latter holds for $x \ge \frac{1+\sqrt{5}}{2}$ which leaves the interval $\big[\sqrt[4]{3}, \frac{1+\sqrt{5}}{2}\big)$ to be checked by hand.

[ EDIT ]  To answer a posted comment, below is the breakdown on the interval $\big[\sqrt[4]{3}, \frac{1+\sqrt{5}}{2}\big)\,$.

$$
\begin{array}{c|lcr}
x \quad & \quad \lfloor x^2\rfloor +\lfloor2x\rfloor\leq\lfloor x^4\rfloor \\
\hline
\big[\sqrt[4]{3}, \,\sqrt{2}\big) \quad & \quad \quad 1\;+\;2\;=\;3 \\
\big[\sqrt{2}, \,\sqrt[4]{5}\big) \quad & \quad \quad 2\;+\;2\;=\;4 \\
\big[\sqrt[4]{5}, \,1.5\big) \quad & \quad \quad 2\;+\;2\;\lt\;5 \\
\big[1.5, \,\sqrt[4]{6}\big) \quad & \quad \quad 2\;+\;3\;=\;5 \\
\big[\sqrt[4]{6}, \frac{1+\sqrt{5}}{2}\big) \quad & \quad \quad 2\;+\;3\;\lt\;6 \\
\end{array}
$$
It follows that the inclusive inequality holds for $x \ge \sqrt[4]{3}\,$, and is a strict inequality for $x \ge \sqrt[4]{6}$.

[ EDIT #2 ] Added the missing break at $\sqrt[4]{5}$ in the table above.
