Calculating value of n

Question

I have this Inequality

$$\displaystyle (n)^{\frac{1}{3}}-(n-1)^\frac{1}{3} \lt \frac{1}{12}$$

I want to calculate the value of n as I have done some calculation I think finding the absolute value is not possible or may be it is possible, I have done some calculation but those calculation does not align with any mathematical concept, I would love to know this particular question belong to which concept of mathematics.
Thanks in advance

Answer 1

We can use the following identity. $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$ and since $$a^2+b^2+c^2-ab-ac-bc=\frac{1}{2}\sum_{cyc}(a-b)^2=0$$ for $a=b=c$ only and $\sqrt[3]{n}\neq-\sqrt[3]{n-1}$, our inequality is equivalent to $$n-(n-1)-\frac{1}{12^3}-3\cdot\frac{1}{12}\sqrt[3]{n(n-1)}<0$$ or $$\sqrt[3]{n^2-n}>4-\frac{1}{432},$$ which gives $n\geq8.508...$ and we have $n\geq9$ in the natural numbers.

Answer 2

1. The function is decreasing if $n\geq1$:

$\begin{align} f(n+1)-f(n)=\\ (n+1)^{\frac{1}{3}}+(n-1)^{\frac{1}{3}} -2n^{\frac{1}{3}}=\\ n^{\frac{1}{3}}\left(\left(1+\frac{1}{n}\right)^{\frac{1}{3}}+\left(1-\frac{1}{n}\right)^{\frac{1}{3}}-2\right)\leq\\ n^{\frac{1}{3}}\left(\left(1+\frac{1}{3n}\right)+\left(1-\frac{1}{3n}\right)-2\right)=0 \end{align}$

2. $g(x):=f(x)-\frac{1}{12}=0 \iff x\sim8.5$
3. Thus, we have $n\geq 9$

Answer 3

You can consider the continuous function $$ f(x)=\sqrt[3]{x}-\sqrt[3]{x-1} $$ over $[1,\infty)$. We have $f(1)=1$; moreover, for $x>1$, $$ f'(x)=\frac{1}{3\sqrt[3]{x^2}}-\frac{1}{3\sqrt[3]{(x-1)^2}}= \frac{\sqrt[3]{(x-1)^2}-\sqrt[3]{x^2}}{3\sqrt[3]{x^2(x-1)^2}} $$ which is obviously negative over $(1,\infty)$, so the function is decreasing.

Now we have $$ f(8)=\sqrt[3]{8}-\sqrt[3]{7}=2-\sqrt[3]{7} $$ and $2-\sqrt[3]{7}>1/12$, because $23^3=12167>7\cdot 12^3=12096$. Also $$ f(9)=\sqrt[3]{9}-\sqrt[3]{8}=\sqrt[3]{9}-2 $$ and $\sqrt[3]{9}-2<1/12$ because $9\cdot 12^3=15552<25^3=15625$.

A pocket calculator is thus sufficient to conclude that the inequality is satisfied by the integers $n\ge 9$.

Why $8$ and $9$? A pocket calculator will show that \begin{align} f(1)-1/12&=11/12\\ f(2)-1/12&\approx0.177\\ f(3)-1/12&\approx0.099\\ f(4)-1/12&\approx0.062\\ f(5)-1/12&\approx0.039\\ f(6)-1/12&\approx0.024\\ f(7)-1/12&\approx0.012\\ f(8)-1/12&\approx0.004\\ f(9)-1/12&\approx-0.003 \end{align} so we know that the root of $f(x)=1/12$ is between $8$ and $9$, but computations with integers are safer than using calculators with floating point operations.

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The other inequality $\sqrt[3]{n}+\sqrt[3]{n-1}<1/12$ is easy as well. The derivative of the associated function is positive on $(0,\infty)$ (except again at $1$), so the function is increasing. Since $\sqrt[3]{0}+\sqrt{0-1}<1/12$ and $\sqrt[3]{1}+\sqrt[3]{1-1}=1>1/12$, the inequality holds only for $n=0$.