For a random graph $G\sim G\left(n,p\right)$ with probability $p$, for every vertex $v$, I need to prove that with high probability, $X=\deg\left(v\right)$ satisfies the condition $\left|X-np\right|\leqslant\sqrt{n\log n}$, at the same time.
For every vertex $v$, I can represent $X$ as a sum of $(n-1)$ indicator random variables $X_{w,v}$ taking 1 if $w$, $v$ are connected, so $\mathbb{E}\left[X\right]=np-p$, and the following can be bounded using Hoeffding's:
$$\mathcal{P}\left(X-np\geqslant\sqrt{n\log n}\right)\leqslant\mathcal{P}\left(X-np+p\geqslant\sqrt{n\log n}\right)=\mathcal{P}\left(\left|X-\mathbb{E}\left[X\right]\right|\geqslant\sqrt{n\log n}\right)\leqslant\frac{2}{n^{2}}$$
But I need to prove it for $\left|X-np\right|$. I tried to break the absolute value into two different terms, one of them is the one from above, and bound each vertex from above, so I will be able to use the union bound to show that they all satisfy it at the same time: $$\mathcal{P}\left(\left|X-np\right|\geqslant\sqrt{n\log n}\right)=\mathcal{P}\left(X-np\geqslant\sqrt{n\log n}\right)+\mathcal{P}\left(-X+np\geqslant\sqrt{n\log n}\right)$$
I thought that I would be able to apply similar technique on the second term, maybe to get it to the form $\left|-X-\mathbb{E}\left[-X\right]\right|$, but I cannot find a way to bound it from above.