Triple Euler sum result $\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$ In the following thread
I arrived at the following result 
$$\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$$
Defining 
$$H_k^{(p)}=\sum_{n=1}^k \frac{1}{n^p},\,\,\, H_k^{(1)}\equiv H_k $$
But, it was after long evaluations and considering many variations of product of polylogarithm integrals.
I think there is an easier approach to get the solution, any ideas ?

We are going to evaluate our sum by establishing a system of two relations. 
Lets establish the first relation and using the derivative of beta function ( see here) , we have
$$-\int_0^1x^{n-1}\ln^3(1-x)\ dx=\frac{H_n^3}{n}+3\frac{H_nH_n^{(2)}}{n}+2\frac{H_n^{(3)}}{n}$$
divide both sides by $n$ then take the sum with respect to $n$, we get
\begin{align}
R_1&=\sum_{n=1}^\infty\frac{H_n^3}{n^2}+3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=-\int_0^1\frac{\ln^3(1-x)}{x}\sum_{n=1}^\infty\frac{x^n}{n}\ dx\\
&=\int_0^1\frac{\ln^4(1-x)}{x}\ dx=\int_0^1\frac{\ln^4x}{1-x}\ dx=24\zeta(5)
\end{align}
Then $$\boxed{R_1=\sum_{n=1}^\infty\frac{H_n^3}{n^2}+3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=24\zeta(5)}$$

To get the second relation, we need to use the sterling number formula ( check here)
$$ \frac{\ln^k(1-x)}{k!}=\sum_{n=k}^\infty(-1)^k \begin{bmatrix} n \\  k \end{bmatrix}\frac{x^n}{n!}$$
letting $k=4$ and using $\displaystyle\begin{bmatrix} n \\  4 \end{bmatrix}=\frac{1}{3!}(n-1)!\left[\left(H_{n-1}\right)^3-3H_{n-1}H_{n-1}^{(2)}+2H_{n-1}^{(3)}\right],$ we get $$\frac14\ln^4(1-x)=\sum_{n=1}^\infty\frac{x^{n+1}}{n+1}\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$
differentiate both sides with respect to $x$, we get
$$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$
multiply both sides by $\ln x/x$ then integrate with respect to $x$, we get
\begin{align}
R_2&=\sum_{n=1}^\infty\frac{H_n^3}{n^2}-3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}\\
&=\int_0^1\frac{\ln^3(1-x)\ln x}{x(1-x)}\ dx=\int_0^1\frac{\ln^3x\ln(1-x)}{x(1-x)}\ dx\\
&=-\sum_{n=1}^\infty H_n\int_0^1x^{n-1}\ln^3x\ dx=6\sum_{n=1}^\infty\frac{H_n}{n^4}=18\zeta(5)-6\zeta(2)\zeta(3)
\end{align}
Then $$\boxed{R_2=\sum_{n=1}^\infty\frac{H_n^3}{n^2}-3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=18\zeta(5)-6\zeta(2)\zeta(3)}$$
now we are ready to calculate our sum:
$$R_1-R_2=6\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}=6\zeta(5)+6\zeta(2)\zeta(3)$$
or 

$$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}=\zeta(5)+\zeta(2)\zeta(3)$$

And as a bonus:
$$R_1+R_2=2\sum_{n=1}^\infty\frac{H_n^3}{n^2}=42\zeta(5)-6\zeta(2)\zeta(3)-4\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}$$
using the definition of $H_n^{(3)}$ and the partial fraction decomposition, its easy to prove $$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\frac{11}2\zeta(5)-2\zeta(2)\zeta(3)$$
which follows 

$$\sum_{n=1}^\infty\frac{H_n^3}{n^2}=10\zeta(5)+\zeta(2)\zeta(3)$$

