Is $O(T+\log T)= O(T\log T)$?
Let $b$ be the base of the logarithm. If $T &gt; \max(b^2, 2)$, then $\log T =\log_b T&gt; 2$. So 
$$T\log T - (T+\log T) = (T-1)(\log T -1) -1 &gt; 1\times 1 -1 =0$$ 
i.e., $T+\log T&lt; T\log T$. 



So, any function that grows asymptotically slower than $T+\log T$ modulo a constant factor also grows asymptotically slower than $T\log T$ modulo the same constant factor. According to the definition of multiple usages of big O-notation, 
$$O(T+\log T)= O(T\log T)$$

Yes, it is true that $O((T+\log T)^{1/n})= O((T\log T)^{1/n})$, where we consider $n$ as a constant.