The recursion $T(n) = T(n/2)+T(n/3)+n$
If $T(1)\le 6$ and $T(2)\le 12$, we can show that $T(n)\le 6n$ by induction.

$$T(n) = T(n/2) + T(n/3) + n \le 6(n/2)+ 6(n/3)+n= 6n.$$ 

More generally, let $c\gt6$ be a constant such that $T(1)\le c$ and $T(2)\le 2c$,  then we can prove routinely that $T(n)\le cn$. Although it might seem unbelievable to you, it is true that
$$T(n)=O(n).$$

Intuitively, since $\dfrac12+\dfrac13\lt1$, the terms $T(n/2)$ and $T(n/3)$ is not big enough to lift $n$ to a power of $n$ of a larger exponent.

There is nothing wrong in your application of Akra-Bazzi method, which tells us that $T(n)=\Theta(n)$.