Here, let $E$ be the set of edges, $I=V_1$ one side of the graph, $J=V_2$ the other side. Let $S$ be the set of edges $(i,j)$ s.t. both $d(i) \geq .9(n/2)$
and $d(j) \geq .9(n/2)$. Further suppose that $|S| \geq .95|E|$ or we are done.

Now let $I_S$ be the set of $i \in I$ s.t. $i$ is incident to an edge in $S$.
Then each $i \in I_S$ has degree at least $.9(n/2)$ in $G$; as there are only $n^2/8$ edges in $G$ it follows that $|I_S|$ is at most $ \frac{m}{.9(n/2)} = \frac{20m}{9n}$. Likewise let $J_S$ be the set of $j$ s.t. $j$ is incident to an edge in $S$. It follows that $|J_S|$ is at most $\frac{20m}{9n}$.

Now, for each $i \in I$, let $d_S(i)$ be the number of edges in $S$ that $i$ is incident to. likewise for each $j \in J$ let $d_S(j)$ be the number of edges in $S$ that $j$ is incident to. Then on the one hand, assuming that $|S| \geq .95|E|$:

$$\sum_{i \in I} d_S(i) = \sum_{i \in I_S} d_S(i) \geq .95 \sum_{i \in I} d(i) \geq .95 \sum_{i \in I_S} d(i)$$

$$\geq .95 |I_s| \left(.9 \times \frac{n}{2}\right) \ \geq \ |I_S| \times \frac{2n}{5}.$$

On the other hand, $\sum_{i \in I_S} d_S(i) \leq |I_S| \times |J_S|$ (since every edge in $S$ goes from $I_S$ to $J_S$), and $J_S$ is no more than $\frac{20m}{9n}$, which is no greater than $n/3$ for $m \leq \frac{n^2}{8}$.

So $S$ cannot be more than $.95|E|$, which as mentioned in the first paragraph, implies your bound.

Extreme values of the sum of squares of degrees of bipartite graphs. Discrete Math. 309, No. 6, 1557-1564 (2009).]. The authors report to have solved your problemcompletely. I did not read the article. Maybe I will, but please do not wait for that. The results are sufficiently complicated to make it impossible (for me at least) to read off the optimal value of $c$. It's also worth pointing out that de Caen's 1998generalbound merely yields a bound of $\frac54 n m$, too large for you. $\endgroup$ – Peter Heinig Mar 21 '18 at 17:45