Showing that $\sqrt[3]{x + \sqrt{\frac{x^3}{6^3} + x^2}} + \sqrt[3]{x - \sqrt{\frac{x^3}{6^3} + x^2}}$ is bounded by 4

Question

I'm trying to show that 4 is an upper bound for $f(x)=\sqrt[3]{x + \sqrt{\frac{x^3}{6^3} + x^2}} + \sqrt[3]{x - \sqrt{\frac{x^3}{6^3} + x^2}}$. This can easily be seen using tools like WolframAlpha. However, I'd like to show this analytically. Any ideas would be appreciated.

Edit:

• The function also has a lower bound of $-12$ but we only care about the upper bound.
• The function is not defined for x < -216 as $\frac{x^3}{6^3} + x^2$ becomes negative for those $x$ values. We are mainly concerned with $x > 0$.
• If we can show that this function is increasing, then using the Taylor expansion of the function at $\infty$ we can show that it is indeed bounded by 4.

Answer 1

Using the identity $(a+b)^3 = a^3 + b^3 + 3ab(a+b)\,$:

$$ \require{cancel} \begin{align} \left(f(x)\right)^3 &= \left(x + \cancel{\sqrt{\frac{x^3}{6^3} + x^2}}\,\right) + \left(x - \cancel{\sqrt{\frac{x^3}{6^3} + x^2}}\,\right) + 3 \sqrt[3]{\bcancel{x^2} - \left(\frac{x^3}{6^3} + \bcancel{x^2}\right)} \, f(x) \\ &= 2 x - \frac{1}{2}\,x\, f(x) \end{align} $$

It is easily verified that $f(x) \gt 0$ on $\mathbb R^+$ and therefore:

$$ \frac{1}{2}\, x\, \big(4 - f(x)\big) = \big(f(x)\big)^3 \;\gt\; 0 \quad\implies\quad f(x) \,\lt\, 4 $$

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Though not used in the above, but since it was brought up both in OP's post and in comments, following is a proof that $f(x)$ is monotonic on $\mathbb R^+$.

Let:

$$ g(x) = f(6x) = \sqrt[3]{\sqrt{x^3 + 36 x^2} + 6x} - \sqrt[3]{\sqrt{x^3 + 36 x^2} - 6x} $$

Then:

$$ \require{cancel} \begin{align} g'(x) = 0 \quad\quad\iff\quad\quad &\frac{x^2 + 24 x + 4 \sqrt{x^3 + 36 x^2}}{\cancel{2 \sqrt{x^3 + 36 x^2}} \sqrt[3]{\sqrt{x^3 + 36 x^2} + 6x}^{\,2}} \\ =\, &\frac{x^2 + 24 x - 4 \sqrt{x^3 + 36 x^2}}{\cancel{2 \sqrt{x^3 + 36 x^2}} \sqrt[3]{\sqrt{x^3 + 36 x^2} - 6x}^{\,2}} \end{align} $$

With $h(x) = \sqrt{x^3 + 36 x^2}\,$:

$$ (x^2 + 24 x + 4 h)^3 (h - 6x)^2 = (x^2 + 24 x - 4 h)^3 (h + 6x)^2 $$

After expanding, collecting and canceling:

$$ \cancel{8 h}\, (16 h^4 + 3 h^2 x^4 - 1152 h^2 x^2 - 3 x^7 - 108 x^6 + 20736 x^4) = 0 $$

Substituting $h^2(x) = x^3 + 36 x^2\,$:

$$ 16 x^6 = 0 $$

It follows that $g'(x)$ does not change sign, so $g(x)$ is monotonic on $\mathbb R^+$.

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[ EDIT ] $\;$ Once either boundedness or monotonicity has been established, passing the equality $\,f(x) = 4 - \frac{2 \left(f(x)\right)^3}{x}\,$ to the limit $x \to\infty$ gives $\,\lim_{x\to\infty} f(x) = 4\,$ so $4$ is a tight upper bound.

Answer 2

It has been proved in the comment section and in the other answer that your sum function $f(x)$ verifies the equation:

$$f(x)^3+\frac x2f(x)-2x=0$$

Interestingly we can (magically$^*$) parametrise this as $\begin{cases}x=\dfrac{2t^3}{(t+16)^2}\\f(x)=\dfrac{4t}{t+16}\end{cases}$

It can easily be derivated and seen that $t\nearrow\implies x\nearrow$ and $f\nearrow$ strictly.

The bounds $t=-12\ (x=-216)$ and $t=+\infty\ (x=+\infty)$ then gives the interval $[-12,4)$ for $f$.

(*) in fact I have been inspired by a similar result I got recently totally unrelated to this problem, not on the net though ( i.e. $x^3+\frac ya x-ay=0$ gives $x=\frac{a^2t}{t+a^4},y=\frac{at^3}{(t+a^4)^2}$ )