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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.

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    $x=1.5$ is counterexample. – 2017-01-08
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    $1.5^2=2.25$, $2(1.5)=3$, and $1.5^4=5.0625$ – 2017-01-08
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    Sorry, it was $1.499^4= 5.049013494001$. – 2017-01-09
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    But the inequality still holds. – 2017-01-09
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    It has ever been a counterexample of dxiv's table or unedited answer. – 2017-01-10
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    @TakahiroWaki The value $x=1.499$ is *not* a counterexample, and never was one in my answer - before or after the edit. Are you maybe confusing the [greatest integer function](http://mathworld.wolfram.com/FloorFunction.html) $\lfloor x \rfloor$ with the [absolute value](https://en.wikipedia.org/wiki/Absolute_value) $|x|$? – 2017-01-10
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    A counterexample against previous table・・・. Now your answer is perfect, though. Well so you say like this, I scored your answer 0. Although no counterexample, nor it was correct answer which satisfies the statement completely. Also so easy? table was not complete. – 2017-01-10
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    @TakahiroWaki Sorry, I don't really follow what you mean to say. I can certainly understand that English is not your most familiar tongue, and of course I don't fault you for that. But it just doesn't parse for me, so I'll leave it at that. – 2017-01-10

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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.

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    $3^{\frac14}< \frac{1+\sqrt5}2$. So check by your hand. – 2017-01-08
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    @TakahiroWaki Not sure what your comment is supposed to mean. If it's about the "*to be checked by hand*" interval, I edited the answer and added the detailed breakdown. – 2017-01-09
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    For example, show $f(x)>3$. "I showed $f(x)>4$, then show between $4>f(x)>3$ by yourself." Is this answer? – 2017-01-09
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    Your answer is wrong strictly. Check $x=1.499$ by wolframalpha or something not your hand. Also I think you didn't verify absolute value of answer. – 2017-01-09
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    @TakahiroWaki The table was indeed missing the break at $\sqrt[4]{5}$ which I added in. For $x \in \big[\sqrt[4]{5},1.5\big)\,$ which includes the case $x=1.499\,$: $\lfloor x^2 \rfloor = 2\,$, $\lfloor 2 x \rfloor = 2\,$ and $\lfloor x^4 \rfloor = 5$ so the inequality holds strictly. – 2017-01-09