Hello bleys

For the last part, can't we consider the differences between $\displaystyle \left(\frac{a}{b}\right)^2$ and $\displaystyle 2$, and $\displaystyle \left(\frac{a+2b}{a+b}\right)^2$ and $\displaystyle 2$?

Which are: $\displaystyle \left|\left(\frac{a}{b}\right)^2 - 2\right|=\left|\frac{a^2-2b^2}{b^2}\right|$

and$\displaystyle \left|\left(\frac{a+2b}{a+b}\right)^2-2\right| =\left|\frac{(a+2b)^2-2(a+b)^2}{(a+b)^2}\right|$$\displaystyle =\left|\frac{2b^2-a^2}{(a+b)^2}\right|$

Clearly the second of these expressions is less than the first. So $\displaystyle \frac{a+2b}{a+b}$ is closer to $\displaystyle \sqrt2$. (Incidentally, doesn't the fact that $\displaystyle a^2-2b^2$ and $\displaystyle 2b^2-a^2$ have opposite signs prove the first part of the question as well?)

Grandad