Let $y=x+1, z=x+2$. Then we must have $\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2} > \frac{1}{45}$. Obviously $\frac{3}{x} > \frac{1}{45} \implies x < 135$. To maximize $x+y+z = 3x+3$, we maximize $x$. If we let $x=134$, then $\frac{1}{134}+\frac{1}{135}+\frac{1}{136} > \frac{1}{45}$, which is true because this is equivalent with

\[\frac{\frac{1}{134}+\frac{1}{136}}{2} > \frac{1}{135}\]

which is true because the inequality

\[\frac{1}{x-1}+\frac{1}{x+1} > \frac{2}{x} \implies \frac{2x}{x^2-1} > \frac{2}{x} \implies x^2>x^2-1\]