Let $ABC$ be the triangle such that $\overline{BC}=a=3$, $\overline{AC}=b=4$ and $\overline{AB}=c=5$. Because $a^2 + b^2 = 3^2 + 4^2 = 5^2 = c^2$, we have that $ABC$ is a right triangle.

Let $M$ be the point inside the triangle such that the distance to his sides is less than $1$. Let $K$, $L$ and $N$ be the points lying on the sides of the triangle. Then, the segments $MK$, $ML$ and $MN$ are the heights to the triangles $AMB$, $BMC$ and $CMA$, respectively. Therefore

$$P_{BMC} + P_{CMA} + P_{AMB} = \frac{xa}{2} + \frac{yb}{2} + \frac{zc}{2} < \frac{a}{2} + \frac{b}{2} + \frac{c}{2} = \frac{3}{2} + \frac{4}{2} + \frac{5}{2} = \frac{12}{2} = 6 = P_{ABC},$$
which is a contradiction with $P_{ABC} = P_{AMB} + P_{BMC} + P_{CMA}$. This means that the point $M$ does not exist.