Prove $\sqrt{2}$ is between $\dfrac{a}{b}$ and $\dfrac{a+2b}{a+b}$ [closed]
Denote $x=\frac{a}{b}$ and $y=\frac{a+2b}{a+b}=\frac{x+2}{x+1}=1+\frac{1}{1+x}$. If $x< \sqrt{2}$ ($x$ cannot be $\sqrt{2}$ because $x$ is a rational), then $1+x< \sqrt{2}+1$, $\frac{1}{1+x}> \frac{1}{\sqrt{2}+1}=\sqrt{2}-1$. Thus $y> \sqrt{2}$.
The other case can be proved similarly.