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 ?

I think this is a shorter solution
By Cauchy product we have 
$$\ln(1-x)\operatorname{Li}_2(x)=-\sum_{n=1}^\infty\left(2\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac3{n^3}\right)x^n$$
multiply both sides by $\frac{\ln(1-x)}{x}$ then integrate from $x=0$ to $x=1$ and use the fact that $-\int_0^1x^{n-1}\ln(1-x)\ dx=\frac{H_n}{n}$ we get
$$\sum_{n=1}^\infty\left(2\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac3{n^3}\right)\left(\frac{H_n}{n}\right)=\int_0^1\frac{\operatorname{Li}_2(x)\ln^2(1-x)}{x}dx=I$$
or $$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}=3\sum_{n=1}^\infty\frac{H_n}{n^4}-2\sum_{n=1}^\infty\frac{H_n^2}{n^3}+I\tag1$$
I proved in this solution
$$\int_0^1\frac{\zeta(2)-\operatorname{Li}_2(x)}{x}\ln^2(1-x)\ dx=2\sum_{n=1}^\infty\frac{H_n^2}{n^3}-2\sum_{n=1}^\infty\frac{H_n}{n^4}$$
or 
$$I=2\zeta(2)\zeta(3)-2\sum_{n=1}^\infty\frac{H_n^2}{n^3}+2\sum_{n=1}^\infty\frac{H_n}{n^4}\tag2$$
plugging $(2)$ in $(1)$ we get
$$ \sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}=5\sum_{n=1}^\infty\frac{H_n}{n^4}-4\sum_{n=1}^\infty\frac{H_n^2}{n^3}+2\zeta(2)\zeta(3)$$
Substituting 
$$\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$
$$\sum_{n=1}^\infty\frac{H_n^2}{n^3}=\frac72\zeta(5)-\zeta(2)\zeta(3)$$
gives
$$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}=\zeta(5)+\zeta(2)\zeta(3)$$
where $\sum_{n=1}^\infty\frac{H_n}{n^4}$ can be obtained from using Euler identity and $\sum_{n=1}^\infty\frac{H_n^2}{n^3}$ can be found here.
