Why if one approximation is too big, the other turns out to be too small, and vice-versa?

Question

$\left|\dfrac{m^2}{n^2}-2\right|$ with $\left|\dfrac{(m+2n)^2}{(m+n)^2}-2\right|$. We need to take absolute values, because if one approximation is too big, the other turns out to be too small, and vice-versa.

Answer 1

Hint:

This amounts to checking that if $m^2<2n^2$, then $$(m+2n)^2>2(m+n)^2 \tag1$$ and similarly for the reverse inequalities.

You just have to expand $(1)$: $$\not m^2+\not4\not m\not n+\not4_{\textstyle2}n^2>\not2m^2+\not4\not m\not n+\not2\not n^2\iff 2n^2>m^2.$$