With the usual notations for the lenghts of the sides of the triangle: $a=5; b=c+3$. $p=\dfrac{a+b+c}{2}=c+4$. $a<b+c\Longrightarrow 5<2c+3\Longrightarrow c>1$. Using the Heron's formula, the area of the triangle is: $A=\sqrt{p(p-a)(p-b)(p-c)}=2\sqrt{(c+4)(c-1)}$. The inradius is: $r=\dfrac{A}{p}=\dfrac{2\sqrt{(c+4)(c-1)}}{c+4}=2\sqrt{\dfrac{c-1}{c+4}}<2$, since $\dfrac{c-1}{c+4}<1$.