We have $$LHS < \int_0^1 \! 1-x^2 \, \mathrm{d}x = \frac{2}{3}.$$

Let $s_i=x_1+...+x_i$. The inequality reduces to $$s_n+s_1s_2^2+...+s_{n-1}s_n^2<s_1^3+...+s_n^3+\frac{2}{3}$$. Now apply AMGM inequalities on $$(1,1,s_n^3),(s_n^3,s_n^3,s_{n-1}^3),...,(s_2^3,s_2^3,s_1^3).$$Summing these gives the desired result.

DerJan wrote: We have $$LHS < \int_0^1 \! 1-x^2 \, \mathrm{d}x = \frac{2}{3}.$$ Can you explain this whole thing more?

Achillys wrote: DerJan wrote: We have $$LHS < \int_0^1 \! 1-x^2 \, \mathrm{d}x = \frac{2}{3}.$$ Can you explain this whole thing more? I think it refers to the area under the curve $y=1-x^2$. Something like this:

Attachments:

Let $a_1,a_2, \ldots ,a_n > 0$ and $k \in \textrm{N}.$ Prove that \[\sum_{i=1}^n a_i\left[1-\left(\sum_{j = 1}^i a_j\right)^k\right] < \frac{k}{k+1}\]

$$x_1\left(1-x_1^2\right)+x_2\left(1-(x_1+x_2)^2\right)+\cdots+x_n\left(1-(x_1+...+x_n)^2\right)\overset{?}{<}\frac{2}{3}.$$\[\frac{2}{3}+\sum{x_k(x_1+...+x_k)^2}\overset{?}{>} x_1+...+x_n\]We have \[\sum{x_k(x_1+...+x_k)^2}=\sum{x_i^3}+2\sum{x_ix_j^2}+\sum{x_i^2x_j}+2\sum{x_ix_jx_k}>\frac{(\sum{x_i})^3}{3}\]where $i<j<k$. \[\frac{(\sum{x_i})^3}{3}+\frac{1}{3}+\frac{1}{3}\overset{\text{AM-GM}}{\geq}\sum{x_i}\]As desired.$\blacksquare$