Is $O(T+\log T)= O(T\log T)$?

Question

Is $O(T+\log T)= O(T\log T)$?

I think this is true but I do not know how to show it mathematically?

Please show it using the definition.

Also, if it is true, is the following true?

$O((T+\log T)^{1/n})= O((T\log T)^{1/n})$

Answer 1

Let $b$ be the base of the logarithm. If $T > \max(b^2, 2)$, then $\log T =\log_b T> 2$. So $$T\log T - (T+\log T) = (T-1)(\log T -1) -1 > 1\times 1 -1 =0$$ i.e., $T+\log T< T\log T$.

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.