how to solve $T (n) = T \left(\frac{n}{2} +\sqrt{n}\right) +\sqrt{6046}$ How can I solve the recurrence 
$$T (n) = T \left(\frac{n}{2} +\sqrt{n}\right) +\sqrt{6046}\ ?$$
Please don't just write the name of the method, as I just started learning this stuff and things are a little hard for me to figure out on my own. 
Thanks in advance.
 A: I believe Akra-Bazzi works for this and gives you $T(x) = \theta(\log x)$.
The recurrence you have satisfies the assumptions of Akra-Bazzi and we get $p=0$ (check out wiki for what $p$ is). $p=0$ implies $T(x) = \theta(\log x)$ in this case (as $g(x)$ is constant).
A more elementary proof would involve showing $T(x) = \mathcal{O}(\log x)$ as in Ross's answer, then try to show $T(x) = \Omega(\log x)$.

Here is an explanation of applying Akra-Bazzi to this one.
Akra-Bazzi is used to solve recurrences of the form:
$$ T(x) = g(x) + \sum_{i=1}^{m} a_i T(b_i x + h_i(x)) \ \text{where}\  x > x_0 $$
where $x$ is the variable, $a_i, b_i$ are constants.
In you case, we have that
$$T(x) = \sqrt{6046} + T(0.5 x + \sqrt{x})$$
Notice that, this corresponds to $\displaystyle m=1$.
The assumptions made by the theorem are
1) $\displaystyle a_i \gt 0$. This is true in your case, as $\displaystyle a_1  = 1$. 
2) $\displaystyle  0 \lt b_i \lt 1$. This is true in your case, as $\displaystyle b_1 = 0.5$. 
3) $\displaystyle |g(x)| = \mathcal{O}(x^c)$ for some $c$. In our case $\displaystyle g(x) = \sqrt{6046}$ is constant. 
4) $h_i(x) = \mathcal{O}(\frac{x}{\log^2 x})$. In our case $h_1(x) = \sqrt{x} = \mathcal{O}(\frac{x}{\log^2 x})$ as $\displaystyle \log^2 x \le K \sqrt{x}$ for sufficiently large $\displaystyle x$.
There are a couple more, but are trivial to verify.
Now to apply Akra-Bazzi theorem, you need to find $\displaystyle p$ such that $\displaystyle \sum_{i=1}^{m} a_i (b_i)^p = 1$.
In our case, we get $\displaystyle 0.5^p = 1$ and so $\displaystyle p = 0$.
Once we find $\displaystyle p$, by the Akra-Bazzi theorem we have that
$$T(x) = \theta\left(x^p \left( 1 + \int_{1}^{x} \frac{g(t)}{t^{p+1}} \ \text{dt}\right)\right)$$
Since $\displaystyle p = 0$ and $\displaystyle g(t)$ is constant, we have that
$$T(x) = \theta\left(1 + \int_{1}^{x} \frac{\sqrt{6046}}{t} \ \text{dt}\right) = \theta(\log x)$$
I would also suggest you try to complete the elementary proof that $\displaystyle T(x) = \Omega(\log x)$. The relation $\displaystyle T(x) \leq T(2x/3) + 80$ only shows that $\displaystyle T(x) = \mathcal{O}(\log x)$. 
The statement $\displaystyle T(x) = \theta(\log x)$ is stronger. For instance, $\displaystyle T(x) = 10$ also satisifies $\displaystyle T(x) \leq T(2x/3) + 80$, but it won't satisfy your original recurrence.
Hope that helps.
A: As you are interested in large $n, \sqrt{n}\lt \frac{n}{a}$ for any $a$.  So you can argue (assuming $T$ is increasing with $n$) that eventually $T(n) \leq T(\frac{2n}{3})+80$.  This fits the method you have been using.
A: You can get precise asymptotics using $f(n)=n/2 \leq n/2 + \sqrt{n} \leq n/2 + \sqrt{n} + 1/2 = g(n)$.  Iterates of $f(n)$ and $g(n)$ have a closed form.
The formulas for $f^k$ and $g^k$ (k'th iterates) imply that the number of iterations of $h(x)=x/2 + \sqrt{x}$ needed to bring $n > 4$ down into any interval $[0,C]$ with $C>4$, is $\log_2{n} + O(1)$.  
From the recurrence, $T(n)$ increases by $\sqrt{6046}$ at each iteration, so the asymptotics are $T(n) = \sqrt{6046} \log_2(n) + O(1)$.
