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

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, $f(3) < f(2)$.
