I have a sketch of proof of irrationality of $\sqrt2$ due to Marcin Mazur. It is asupposed that $\sqrt2=\frac ab$ where $a$ is the smallest positive numerator that $\sqrt 2$ can have as a fraction. Since $\sqrt2=\frac{-4\sqrt2+6}{3\sqrt2-4}$, we have $\sqrt 2=\frac{4a-6b}{4b-3a}$. In order to get a contradiction we use the fact that $1<\sqrt2=\frac ab<2$ or $b<a<2b$. Now, $a<2b$ implies that

$$3a<6b\implies4a<6b+a\implies 4a-6b<a$$

I've tried a lot to prove that $0<4a-6b$, but couldn't arrive to any result. Could you help me, please?