50. Let $a, b, c, d, e$ be positive real numbers, and $a^{3}+a b+b^{3}=c+d=1$, prove: $\sum_{a c}\left(a+\frac{1}{a}\right)^{3} \geqslant$
40. (2003 Greek Mathematical Olympiad Problem)

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50. Since $c+d=1$, by the AM-GM inequality $(c+d)\left(\frac{l}{c}+\frac{1}{d}\right) \geqslant 4$, and by the power mean inequality we get
$$\begin{array}{l}
\left(c+\frac{1}{c}\right)^{3}+\left(d+\frac{1}{d}\right)^{3} \geqslant \frac{1}{4}\left[\left(c+\frac{1}{c}\right)+\left(d+\frac{1}{d}\right)^{2}\right]^{3}= \\
\frac{1}{4}\left[(c+d)+\left(\frac{1}{c}+\frac{1}{d}\right)^{3} \geqslant \frac{125}{4}\right.
\end{array}$$

Assume without loss of generality that $a=\min \{a, b\}$, then $1=a^{3}+a b+b^{3} \geqslant 2 a^{3}+a^{2}$, so $a \leqslant \frac{2}{3}$, otherwise, if $a \geqslant \frac{2}{3}$, then
$$a^{3}+a b+b^{3} \geqslant 2 a^{3}+a^{2} \geqslant 2\left(\frac{2}{3}\right)^{3}+\left(\frac{2}{3}\right)^{2}=\frac{28}{27}>1$$

Therefore, $a \leqslant \frac{2}{3}$. Since the function $y=x+\frac{1}{x}$ is monotonically decreasing on $(0,1)$, when $a \leqslant \frac{2}{3}$, $a+\frac{1}{a} \geqslant \frac{2}{3}+\frac{3}{2}=\frac{13}{6}$, thus
$$\left(a+\frac{1}{a}\right)^{3} \geqslant\left(\frac{13}{6}\right)^{3} \geqslant \frac{2197}{216}>10>\frac{35}{4}$$

Therefore,
$$\sum_{c x}\left(a+\frac{1}{a}\right)^{3} \rightarrow\left(a+\frac{1}{a}\right)^{3}+\left(c+\frac{1}{c}\right)^{3}+\left(d+\frac{1}{d}\right)^{3} \geqslant 40$$