Here is an answer to the first version of the post, now modified:
$$\max(\xi^2,\eta^2)\leqslant\xi^2+\eta^2\implies\mathrm E(\max(\xi^2,\eta^2))\leqslant\mathrm E(\xi^2)+\mathrm E(\eta^2)=2\leqslant1+\sqrt{1+r^2}$$
Here are some detailed hints leading to a solution of the revised version.
First, $\max(\xi^2,\eta^2)=\frac12(\xi^2+\eta^2)+\frac12|\eta^2-\xi^2|$ hence, using the notation $s=\sqrt{1-r^2}$, one sees that the desired inequality is equivalent to $\mathrm E|\eta^2-\xi^2|\leqslant2s$. Second, there exists a random variable $\zeta$ such that $\mathrm E(\zeta)=\mathrm E(\zeta\xi)=0$, $\mathrm E(\zeta^2)=1$ and $\eta=r\xi+s\zeta$. Hence, one wants to show that $\mathrm E|\kappa|\leqslant2$, where
$$
\kappa=s\zeta^2+2r\xi\zeta-s\xi^2.
$$
If, by chance, the pointwise inequality
$$
-\xi^2-\zeta^2\leqslant\kappa\leqslant\xi^2+\zeta^2\tag{$\ast$}
$$
holds, the proof is over. Now, remember that $s=\sqrt{1-r^2}$ and show that $(\ast)$ holds.