Q:

Big O Notation Arithmetics

Assuming it is known that:
$$ t\sqrt n  = O((nt)^{2/3}+n+t)$$
How is it possible to deduce a bound on $ t $?
Specifically, I need to prove that $ t = O(\sqrt n)$. This is part of a bigger question, so I am not sure the above formula is correct. I would like to know in general how to solve such problems.

A:

Assume that $t\sqrt{n}\leqslant c\cdot((nt)^{2/3}+n+t)$ for every $n\geqslant n_0$. 
Choose $n_0$ large enough for $n_0\geqslant4c^2$ to hold. Then, $ct\leqslant\frac12t\sqrt{n}$ for every $n\geqslant n_0$, hence $t\sqrt{n}\leqslant2c\cdot((nt)^{2/3}+n)$, that is $x\leqslant2c\cdot(x^{2/3}+1)$, where $x=t/\sqrt{n}$. 
Since $z-2c\cdot(z^{2/3}+1)\to+\infty$ when $z\to+\infty$, $z-2c\cdot(z^{2/3}+1)\gt0$ for every $z$ large enough, say $z\gt K_c$. This means that $x\leqslant K_c$, that is, that $t\leqslant K_c\sqrt{n}$ for every $n\geqslant n_0$.

