<s>[INST] A cubic nonlinear Euler sum Any idea how to solve the following Euler sum 

$$\sum_{n=1}^\infty \left( \frac{H_n}{n+1}\right)^3 =
 -\frac{33}{16}\zeta(6)+2\zeta(3)^2$$

I think It can be solved it using contour integration but I am interested in solutions using real methods. 
 [/INST] \begin{align}
S&=\sum_{n=1}^\infty\left(\frac{H_n}{n+1}\right)^3=\sum_{n=1}^\infty\left(\frac{H_{n-1}}{n}\right)^3\\
&=\sum_{n=1}^\infty\frac{H_n^3}{n^3}-3\sum_{n=1}^\infty\frac{H_n^2}{n^4}+3\sum_{n=1}^\infty
\frac{H_n}{n^5}-\sum_{n=1}^\infty\frac{1}{n^6}
\end{align}
Substituting the following results:
$$\sum_{n=1}^\infty\frac{H_n^3}{n^3}=\frac{93}{16}\zeta(6)-\frac52\zeta^2(3)$$
$$\sum_{n=1}^\infty \frac{H_n^2}{n^4}=\frac{97}{24}\zeta(6)-2\zeta^2(3)$$
$$\sum_{n=1}^\infty\frac{H_n}{n^5}=\frac74\zeta(6)-\frac12\zeta^2(3)$$
We get $$\boxed{S=2\zeta^2(3)-\frac{33}{16}\zeta(6)}$$
Note that the first and second sum are proved here and here respectively. As for the third sum, can be obtained using the Euler identity.
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