Analytical solution:
Let f(x_1,x_2,x_3,x_4) = x_1^3+x_2^3+x_3^3+x_4^3-(x_1+x_2+x_3+x_4), and g(x_1,x_2,x_3,x_4) = x_1x_2x_3x_4-1.
we will find the minimum of f under the condition g = 0.
We find the solutions to the lagrange equations
\frac{\delta f}{\delta x_i} = \lambda \frac{\delta g}{\delta x_i}
and g = 0, or
3x_i^2 -1= \lambda x_1x_2x_3 \Leftrightarrow 3x_i^3-x_i=\lambda, so 3x_i^3-x_i = 3x_j^3-x_j, or (x_i-x_j)(3(x_i^2+x_ix_j+x_j^2)-1) = 0. Pick the largest one, say x_k. It will be larger or equal than 1. And thus 3(x_k^2+x_ix_k+x_k^2)-1 \geq 2 > 0. Hence x_i = x_k for all i, so they are all equal. Hence x_i = 1 for all i.
Now this is the minimum on some compact set containing (1,1,1,1).
The set determined by x_i > 0, and x_1x_2x_3x_4 = 1, and x_1+x_2+x_3+x_4 <=5 contains (1,1,1,1) and is compact.
Furthermore, by the general QM-AM \sqrt[3]{\frac{x_1^3+x_2^3+x_3^3+x_4^3}{4}} \geq \frac{x_1+x_2+x_3+x_4}{4}
So x_1^3+x_2^3+x_3^3+x_4^3 \geq \frac{(x_1+x_2+x_3+x_4)^2}{16} (x_1+x_2+x_3+x_4)
Hence outside our set f(x_1,x_2,x_3,x_4) > 0. So (1,1,1,1) is actually the value for which f is minimized, and the value is 0. Hence x_1^3+x_2^3+x_2^3+x_4^3 \geq x_1+x_2+x_3+x_4.
(EDIT: I assumed we were working in the subset x_i > 0 originally, so just assume I said that before I did :) )
