Let $X\in\{0,1\}^2$ have mean $\mu=\left[\begin{smallmatrix}p\_1\\p\_2\end{smallmatrix}\right]$ and $\Pr[X\_1 = X\_2 = 1] = p\le \min\{p\_1,p\_2\}$.
(Note we must have $1-p\_1-p\_2+p\ge 0$ for the distribution to be well defined.)
We can then compute the covariance matrix $\Sigma = E[(X-\mu)(X-\mu)^T] = \left[\begin{smallmatrix}p\_1(1-p\_1)&p-p\_1p\_2\\p-p\_1p\_2&p\_2(1-p\_2)\end{smallmatrix}\right]$.


I would like to use the Berry Essen bound, and for that we need to upper bound the quantity $\gamma=E[\|\Sigma^{-1/2}(X-\mu)\|\_2^3]$.


I believe one should be able to show
$$\gamma \le C\left(\tfrac1{\sqrt{p\_1(1-p\_1)}}+\tfrac1{\sqrt{p\_2(1-p\_2)}}+\tfrac1{\sqrt{\min\{p\_1,p\_2\}-p}}\right)
, 
$$
for some universal constant $C>0$.


The symbolic computation of $\Sigma^{-1/2}$ is a bit unwieldy though, and so I wonder if there is some tricks I may use to arrive at this result more neatly?


Or if not, any proof would be appreciated.


**Update:** Using Pinelli's observation, we can compute
$$\begin{align}\gamma |\Sigma|^{3/2}
&= p[(1-p\_1)(1-p\_2)v]^{3/2}
\\&+ (p\_1-p)[(1-p\_1)p\_2(1-v)]^{3/2}
\\&+ (p\_2-p)[p\_1(1-p\_2)(1-v)]^{3/2}
\\&+ (1-p\_1-p\_2+p)[p\_1 p\_2 v]^{3/2}
,
\end{align}$$
where $v = p\_1+p\_2-2p$ and $|\Sigma|=p\_1p\_2(1-v)-p^2$.


It seems like I forgot a factor $1/\sqrt{p}$.
In fact it seems that bounds of $\gamma \le
\frac{2}
{\sqrt{p\,(\min\{p\_1,p\_2\}-p)}}
$ or $\gamma \le
\frac{1}
{\sqrt{p}}+
\frac{2}
{\sqrt{\min\{p\_1,p\_2\}-p}}
$ are sufficient and correct.



Bounding $E[\|\Sigma^{-1/2}(X-\mu)\|_2^3]$ for 2-dimensional Bernoulli
Your inequality does not hold in general:


You don't need to compute to compute $\Sigma^{-1/2}$, because $\|\Sigma^{-1/2} x\|\_2^2=x^T\Sigma^{-1}x$ for all $x\in\mathbb R^2$. Using this simple observation with e.g. $p=0$, $p\_1=1/2$, $p\_2=1/2-\epsilon$, and $\epsilon\downarrow0$, we find 
$$\gamma
=\frac{\left(1-\epsilon -\epsilon ^2\right)^{3/2}}{\sqrt{\epsilon }}+\frac{1}{2} (1+\epsilon
 )^{3/2}+\frac{\left(1-\epsilon +2 \epsilon ^2\right)^{3/2}}{2 \sqrt{1-2 \epsilon }}\to\infty,$$
whereas 
$$\frac1{\sqrt{p\_1(1-p\_1)}}+\frac1{\sqrt{p\_2(1-p\_2)}}+\frac1{\sqrt{\min(p\_1,p\_2)-p}}\to4+\sqrt2.$$


