<s>[INST] One-variable inequality Let $t \in [0;1]$. Prove that
$$\sqrt[3]{{t{{\left( {t + 1} \right)}^2} + 4\left( {t + 1} \right)}} - \sqrt[3]{{12t}} \ge \frac{{\sqrt[3]{{14}}}}{7}\left( {\sqrt[3]{{2{{(t + 7)}^2}(3t + 1)}} - 8\sqrt[3]{t}} \right).$$
Direction 1: I consider two functions $f,g$ as follows
$$f\left( t \right) = \sqrt[3]{{t{{\left( {t + 1} \right)}^2} + 4\left( {t + 1} \right)}} - \sqrt[3]{{12t}},g\left( t \right) = \frac{{\sqrt[3]{{14}}}}{7}\left( {\sqrt[3]{{2{{(t + 7)}^2}(3t + 1)}} - 8\sqrt[3]{t}} \right)$$
However both functions $f,g$ reduce the function (while what I want is for them to be opposite monotonous).
Direction 2: I transform the inequality as follows
$$\begin{array}{l}
{(t - 1)^2}(t + 4) + 3\sqrt[3]{{t{{\left( {t + 1} \right)}^2} + 4\left( {t + 1} \right)}} \cdot \sqrt[3]{{12t}}\left( {\sqrt[3]{{t{{\left( {t + 1} \right)}^2} + 4\left( {t + 1} \right)}} - \sqrt[3]{{12t}}} \right)\\
 \ge \frac{4}{{49}}{(t - 1)^2}(3t + 49) + 3\sqrt[3]{{2{{(t + 7)}^2}(3t + 1)}} \cdot 8\sqrt[3]{t}\left( {\sqrt[3]{{2{{(t + 7)}^2}(3t + 1)}} - 8\sqrt[3]{t}} \right).
\end{array}$$
It was too complicated and I stopped.
I have also submitted this question to the AoPS forum at the following link, but no support response yet!
https://artofproblemsolving.com/community/u797276h3001638p26953179
 [/INST] The desired inequality is written as
$$\sqrt[3]{a} - \sqrt[3]{b} \ge \sqrt[3]{c} - \sqrt[3]{d}$$
or
$$\frac{a - b}{a^{2/3} + a^{1/3}b^{1/3} + b^{2/3}}
\ge \frac{c - d}{c^{2/3} + c^{1/3}d^{1/3} + d^{2/3}} \tag{1}$$
where
$a = t(t + 1)^2 + 4(t + 1)$,
$b = 12t$,
$c = \frac{14}{7^3}\cdot 2(t + 7)^2(3t + 1)$, and
$d = \frac{14}{7^3}\cdot 8^3t$.
We have
\begin{align*}
 a - b &= (t + 4)(t - 1)^2 \ge 0, \\
c - d &= \frac{4}{49}(3t + 49)(t - 1)^2 \ge 0,\\
 (a - b) - (c - d) &= \frac{37}{49}t(t-1)^2 \ge 0,\\
  c - a &= \frac{t(-37t^2+74t+399)}{49} \ge 0, \\
 d - b &= \frac{436t}{49} \ge 0.
\end{align*}
Thus, we have
$c \ge a > 0$ and $d \ge b \ge 0$ which results in
$$a^{2/3} + a^{1/3}b^{1/3} + b^{2/3} \le c^{2/3} + c^{1/3}d^{1/3} + d^{2/3}.$$
Also, we have $a - b \ge c - d \ge 0$.
Thus, (1) is true.
We are done.
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