We use strong induction, also known sometimes as course of values induction. This version of mathematical induction comes up very often in Computer Science arguments (and elsewhere, like Number Theory). Any strong induction argument can be mechanically recast as an ordinary induction argument, but the strong induction approach is often more natural.
Assume that we know, for a specific natural number $n$, that $T(k+1)>T(k)$ for every natural number $k< n$. We show that $T(k+1)>T(k)$ for every natural number $k < n+1$.
Obviously we only need to prove that $T(n+1)>T(n)$. This almost follows from the induction hypothesis, since $T(n+1)=T((n+1)/2+\sqrt{n+1})+\sqrt{6046}$ and $T(n)=T(n/2+\sqrt{n})+\sqrt{6046}$, and by the induction hypothesis, $T((n+1)/2+\sqrt{n+1})>T(n/2+\sqrt{n})$.
But we must be able to assume that $(n+1)/2+\sqrt{n+1}>n/2+\sqrt{n}$. This last inequality is obviously true in the reals. However, if we are going to do rounding, and "$k/2+\sqrt{k}$" is not really $k/2+\sqrt{k}$, all bets are off, and we can only conclude that $T(n+1)\ge T(n)$. There may not be a change when we increment $n$ by $1$, but if there isn't, there will be a change when we increment $n$ once more.
Remark: Let's illustrate the rounding issue numerically. Let $n=50$. Then $n/2+\sqrt{n}\approx 18.1$ and $(n+1)/2+\sqrt{n+1}\approx 18.7$. If we use the greatest integer function, these both become $18$, so $T(26)=T(27)=T(18)$. It is easy to come up with a similar example for which using the nearest integer function has similar issues.