Rearranging asymptotic notation

Question

If $a \le b^{\frac{1+\log_{2}b}{2}}(1+o(1))$, then what is $b$ in terms of $a$? Whenever I try to rearrange this, I get in a huge mess...

Any help would be appreciated. Thanks.

Answer 1

From what I can see (and if I didn't make any mistake), it should be something like \begin{align*} b\geq 2^{\displaystyle\sqrt{2\log_2 a }-\dfrac{1}{2}+\dfrac{1}{8\sqrt{2\log_2a}} + o\left(\frac{1}{\sqrt{\log_2a}}\right)}. \end{align*} Plugging in back the expression, you get $$ a \leq a\cdot 2^{\frac{1}{256\log_2 a}}\cdot\left(1+o(1)\right) = a \left(1+o(1)\right) $$ when $a\to\infty$.