1 - $\lim\langle Au_n,u_n-u\rangle=0$
Indeed, ofr $w=u$ we have
\begin{eqnarray} 0 &\leq& \liminf\langle Au_n,u_n-u\rangle \nonumber \\ &\leq& \limsup\langle Au_n,u_n-u\rangle \nonumber \\ &\le& 0 \end{eqnarray}
Therefore, $\lim\langle Au_n,u_n-u\rangle=0$
2 - $\langle Au,u-w\rangle\leq\liminf\langle Au_n,u-w\rangle,\ \forall\ w\in X$
We have, (im using 1 here)
\begin{eqnarray} \langle Au,u-w\rangle &\leq& \liminf\langle Au_n,u_n-w\rangle+\liminf\langle Au_n,-u_n+u\rangle \nonumber \\ &\leq& \liminf\langle Au_n,u_n-u_n+u-w\rangle \nonumber \\ &=& \liminf\langle Au_n,u-w\rangle \end{eqnarray}
3 - $\langle Au_n,v\rangle\rightarrow\langle Au,v\rangle,\ \forall\ v\in X$
Take $w=u-v$ in 2. Hence
\begin{eqnarray} \langle Au,v\rangle &\leq& \liminf\langle Au_n,v\rangle \nonumber \\ &\leq& \limsup\langle Au_n,u_n-u_n+u-u+v\rangle \nonumber \\ &\le& \limsup\langle Au_n,u_n-u\rangle+\limsup\langle Au_n,-u_n+u+v\rangle \\ &=& \limsup-\langle Au_n,u_n-(u+v)\rangle \\ &=& -\liminf\langle Au_n,u_n-(u+v)\rangle \\ &\leq& -\langle Au,u-(u+v)\rangle \\ &=& \langle Au,v\rangle \end{eqnarray}
Because $X$ is reflexive, by using 3 we get that $Au_n\rightharpoonup Au$
4 - $\lim\langle Au_n-Au,u_n-u\rangle\rightarrow 0$
By using the hypothesis with $w=u$ we have that
\begin{eqnarray} 0 &\leq& \liminf\langle Au_n,u_n-u\rangle \nonumber \\ &=& \liminf\langle Au_n,u_n-u\rangle+\liminf\langle Au,u_n-u\rangle \nonumber \\ &\le& \liminf\langle Au_n-Au,u_n-u\rangle \\ &\leq& \limsup\langle Au_n-Au,u_n-u\rangle \\ &\leq& \limsup\langle Au_n,u_n-u\rangle \\ &\leq& 0 \end{eqnarray}
5 - $\langle Au_n,u_n\rangle\rightarrow\langle Au,u\rangle$
Note that $\langle Au_n,u_n\rangle=\langle Au_n-Au,u_n-u\rangle-\langle Au,u\rangle+\langle Au_n,u\rangle+\langle Au,u_n\rangle$
It follows from 3,5 and $u_n\rightharpoonup u$ that $\lim\langle Au_n,u_n\rangle=\langle Au,u\rangle$
Please verify if my proof is correct.
Id like to observe that the converse of this theorem is also true and much more easy to prove.