Prove $\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$

Question

I am a student in Germany, and I prepare for Math Olympiad by solving math problems. I have been solving the following question, which took about 4 hours to solve.

Prove the following inequality without using calculator: $$\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$$

Can you check my proof? It would be really grateful.

First, we can define function $f(x)$ as following: $$f(x) = \sqrt[3]{x+1} - \sqrt[3]{x}\space(x > 0)$$ $$f(3) = \sqrt[3]{3+1} - \sqrt[3]{3} = \sqrt[3]{4} - \sqrt[3]{3}$$ $$f(2) = \sqrt[3]{2+1} - \sqrt[3]{2} = \sqrt[3]{3} - \sqrt[3]{2}$$

Then, we will differentiate $f(x)$ to check whether $f(x)$ is a decreasing function or not. $f '(x)$ must be a falling function if $f '(x)$ < 0. $$f'(x) = \frac{1}{3\sqrt[3]{(x+1)^2}} - \frac{1}{3\sqrt[3]{x^2}}$$ Since the minuend is smaller than the subtrahend (minuend has a bigger denominator than the denominator of subtrahend), we can say $f '(x)$ is less than 0 which makes $f(x)$ a decreasing function. Falling function means that $f(a) > f(a+1)$. Substitute $a=2$ and we get: $$f(2) > f(3)$$ $$\sqrt[3]{3} - \sqrt[3]{2} > \sqrt[3]{4} - \sqrt[3]{3}$$ $$\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$$

Thank you for reading this text, but it would be more grateful if you check my solution, and comment my solution.

I wish you a beautiful day, and stay safe.

Answer 1

Your solution is fine. If you want a non-calculus approach, note that \begin{align*} f(x) &= (x+1)^{1/3}-x^{1/3} \\ &= \dfrac{\left((x+1)^{1/3}-x^{1/3}\right)\left((x+1)^{2/3}+(x+1)^{1/3}x^{1/3}+x^{2/3}\right)}{(x+1)^{2/3}+2(x+1)^{1/3}x^{1/3}+x^{2/3}} \\ &= \dfrac{(x+1)-x}{(x+1)^{2/3}+(x+1)^{1/3}x^{1/3}+x^{2/3}} \\ &= \dfrac{1}{(x+1)^{2/3}+(x+1)^{1/3}x^{1/3}+x^{2/3}} \end{align*} is clearly decreasing since the numerator is constant and positive while the denominator is increasing and positive. Hence, $f(3) < f(2)$.

Answer 2

I guess your question has been answered. Here is a different way to see it. Note that one has $$a^3-b^3=(a-b)(a^2+ab+b^2).$$ It follows that $$1=(\sqrt[3]{4}-\sqrt[3]{3})(\sqrt[3]{16}+\sqrt[3]{12}+\sqrt[3]{9})=(\sqrt[3]{3}-\sqrt[3]{2})(\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4}).$$ It suffices now to show that $$\sqrt[3]{16}+\sqrt[3]{12}+\sqrt[3]{9}>\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4}$$

$$\Leftrightarrow \sqrt[3]{16}+\sqrt[3]{12}>\sqrt[3]{6}+\sqrt[3]{4},$$ which is clear.

Answer 3

Alternative approach

No calculators allowed.
Requires knowledge that
$\log_{10} (2) \approx 0.301$ and $\log_{10} (3) \approx 0.477.$

The problem reduces to showing that $(4)^{(1/3)} + (2)^{(1/3)} < 2(3)^{(1/3)}.$

Cubing both sides, this resolves to showing that
$4 + 2 + 3\left[4^{(2/3)}2^{(1/3)} + 4^{(1/3)}2^{(2/3)}\right] < 24.$

This in turn resolves to showing that
$4^{(2/3)}2^{(1/3)} + 4^{(1/3)}2^{(2/3)} < 6.$

Converting to logarithms, base $(10)$,
this resolves to showing that
$10^{(1.505/3)} + 10^{(1.204/3)} < 6.$

This resolves to showing that

$$10^{(0.5017)} + 10^{(0.4014)} < 6.\tag 1$$

$\log_{10}(3.2) \approx (5 \times 0.301) - 1 = 0.505.$
Therefore, $10^{(0.5017)} < 3.2.$

Further, $\log_{10} (2.7) \approx (0.477 \times 3) - 1 = 0.431.$
Therefore $10^{(0.4014)} < 2.7.$

Therefore, the LHS of expression (1) above is $< 5.9$.

Answer 4

Consider also a simpler function $g(x) = \sqrt[3]x$. For $x > 0$, $g'(x) = \dfrac1{3x^{2/3}}$ is decreasing.

By mean value theorem, there exists some $c_1 \in (2,3)$ and some $c_2 \in (3,4)$ that satisfy

$$g'(c_1) = \frac{\sqrt[3]3-\sqrt[3]2}{1};\quad g'(c_2) = \frac{\sqrt[3]4-\sqrt[3]3}{1}$$

And since $c_2 > c_1$ and so $g'(c_2) < g'(c_1)$,

$$\sqrt[3]4-\sqrt[3]3 < \sqrt[3]3-\sqrt[3]2$$

Answer 5

The standard difference quotient for approximating $f''(a)$ is $$ \frac{f(a-h) - 2 f(a) + f(a+h)}{h^2} $$ Here $f(x) = \sqrt[3]x $ is infinitely differentiable, while the second and fourth derivatives are negative for $0 < h < a.$ Indeed $$ f(a-h) - 2 f(a) + f(a+h) = f''(a) h^2 + \frac{h^3}{6} \left( f'''( \xi) - f'''(\eta) \right) $$

where $ a-h < \eta < a < \xi < a+h .$ This is just the Taylor series with remainder. As the fourth derivative is negative we see $ f'''( \xi) - f'''(\eta) < 0.$ Thus we find, $$ f(a-h) - 2 f(a) + f(a+h) <0$$

Her $f$ is the cube root, $a=3$ and $h=1$ so $$ \sqrt[3]2 - 2\sqrt[3]3 + \sqrt[3]4 < 0 $$

Answer 6

Your proof is right.

I think, there is something better.

We can use the following identity. $$a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)\sum_{cyc}(a-b)^2.$$ We obtain: $$ \sqrt[3]2 - 2\sqrt[3]3 + \sqrt[3]4 < 0 $$ it's $$2-24+4+3\cdot\sqrt[3]2\cdot2\sqrt[3]3\cdot\sqrt[3]4 < 0 $$ or $$12\sqrt[3]3<18$$ or $$24<27,$$ which is true.

Id est, $$\sqrt[3]2 - 2\sqrt[3]3 + \sqrt[3]4 < 0$$ is true.