<s>[INST] 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.
 [/INST] 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 $< 5.9$.
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