So I want to prove that the recurrence $H(n)=\sqrt{n}H(\sqrt{n})+\log_2 n$ is bounded above by $O(n)$ or there exists $ c\in \mathbb{R}^{+}$:
$$H(n)=\sqrt{n}H(\sqrt{n})+\log_2 n\leq cn$$
Now I couldn't get a $O(n)$ bound, as subbing in $cn$ I get $H(n)=\sqrt{n}(c\sqrt{n})+\log_2 n$, which is bigger than $cn$. Thus I tried to get rid of the end term by setting my induction hypothesis as $c_1n-c_2\log_2 n$ which is stronger, which by subbing it into my recurrence I got:
\begin{align*} \sqrt{n}(c_1\sqrt{n}-c_2\log_2 \sqrt{n})+\log_2 n&=c_1 n-\sqrt{n}c_2\log_2 \sqrt{n}+\log_2 n \\ \text{ (Now I will establish it again our guessed upper bound)}\\ c_1 n-\sqrt{n}c_2\log_2 \sqrt{n}+\log_2 n&\leq c_1 n -c_2 \log_2 n\\ -\sqrt{n}c_2\log_2 \sqrt{n}+\log_2 n&\leq -c_2 \log_2 n\\ (1/2)\sqrt{n}c_2\log_2 n+\log_2 n&\geq c_2 \log_2 n c_2\geq \frac{2}{\sqrt{n}-2} \end{align*}
I need to find a constant $c_2$ that satisfies the inequality above (as I have $c_2$ in terms of $n$ or bound $(1/2)\sqrt{n}\log_2 n$ by something of that same order. Which I am stuck on.