4. Find the largest constant $C$ such that
$$
\sum_{i=1}^{4}\left(x_{i}+\frac{1}{x_{i}}\right)^{3} \geq C
$$
for all positive real numbers $x_{1}, \cdots, x_{4}$ such that
$$
x_{1}^{3}+x_{3}^{3}+3 x_{1} x_{3}=x_{2}+x_{4}=1 \text {. }
$$
4. First note that
$$
\begin{aligned}
x_{1}^{3}+x_{3}^{3}+3 x_{1} x_{3}-1 & =x_{1}^{3}+x_{3}^{3}-(1)^{3}-3 x_{1} x_{3}(-1) \\
& =\left(x_{1}+x_{3}-1\right)\left(\left(x_{1}+x_{3}\right)^{2}+\left(x_{1}+x_{3}\right)+1\right)-3 x_{1} x_{3}\left(x_{1}+x_{3}-1\right) \\
& =\left(x_{1}+x_{3}-1\right)\left[\left(x_{1}-x_{3}\right)^{2}+\left(x_{1}+1\right)\left(x_{3}+1\right)\right] .
\end{aligned}
$$

It is equal to zero only when either $x_{1}+x_{3}=1$ or $x_{1}=x_{3}=-1$. Thus we must have $x_{1}+x_{3}=1$ as they are positive. It now suffixes to show that the following is sharp:
$$
\sum_{i=1}^{2}\left(y_{i}+\frac{1}{y_{i}}\right)^{3} \geq 125 / 4 \text { when } y_{1}+y_{2}=1 \text { and } y_{1}, y_{2}>0 \text {. }
$$

To this end, it is clear that the function $f(x)=(x+1 / x)^{3}$ is convex. Thus,
$$
f(x)+f(1-x) \geq 2 f(1 / 2)=125 / 4 .
$$