OABC is a tetrahedron of volume $\frac{2}{3}$ cubic unit. Also $BC=\sqrt{2},\angle AOB=\frac{\pi }{4}$ and OA + OB = 4, then which is/are CORRECT?

$$\begin{gathered} \because OA+OB=4 \\ \Rightarrow OA.OB\le 4 \end{gathered}$$

$\therefore ar\left(\Delta AOB\right)=\frac{1}{2}\left(OA\right)\left(OB\right)\times \frac{1}{\sqrt{2}}\le \sqrt{2}$

Let h = altitude drawn from the vertex C to the base AOB.
So,  $h\le \sqrt{2}$
 $\therefore V=\frac{1}{3}\times ar\left(AOB\right)\times h\le \frac{1}{3}\times \sqrt{2}\times \sqrt{2}=\frac{2}{3}$
But $V=\frac{2}{3}\Rightarrow$  equality holds everywhere
So, $OA = OB = 2 \& h=\sqrt{2}$
Also, BC $\perp$ base  $AOB\Rightarrow \angle OBC={90}^0$
So,  $OC=\sqrt{6}$
And,  $AB=\sqrt{4+4-8\times \frac{1}{\sqrt{2}}}=\sqrt{8-4\sqrt{2}}$
$$\begin{gathered} \therefore AC=\sqrt{8-4\sqrt{2}+2}=\sqrt{10-4\sqrt{2}} \\ >\sqrt{9-4\sqrt{2}}=(2\sqrt{2}-1) \end{gathered}$$