2018 January Challenge: Prove inequality in geometry problem

Question

Considering it's the beginning of a new year, I have created the following challenge. I hope to make one every month until December 2018!

Here goes:

Show that $AD-AB>AC^3$. Do not use scale drawings.

Answer 1

We need to prove that $$\sqrt{1-\frac{1}{\sqrt2}}-\sqrt{\frac{3}{2}}(\sqrt2-1)>\left(1-\frac{1}{\sqrt2}\right)^3,$$ which is true.

Indeed, let $DFMK$ be the given rectangle,

$\Delta FBM$ be given triangle ($FM=BM=1$,$\measuredangle FMB=90^{\circ}$) and

$\Delta EFC$ be another given triangle ($EF=FC=1$, $\measuredangle EFC=90^{\circ}$).

Thus, $$\frac{AC}{EF}=\frac{BC}{BF}$$ or $$\frac{AC}{1}=\frac{\sqrt2-1}{\sqrt2},$$ which gives $$AC=1-\frac{1}{\sqrt2}.$$ Hence, $$AB=\sqrt{AC^2+BC^2}=\sqrt{\left(1-\frac{1}{\sqrt2}\right)^2+(\sqrt2-1)^2}=\sqrt{\frac{3}{2}}(\sqrt2-1).$$ Now, since $\measuredangle DCA=45^{\circ}$, by law of cosines for $\Delta DAC$ we obtain: $$\begin{align}AD&=\sqrt{\left(\frac{1}{\sqrt2}\right)^2+\left(1-\frac{1}{\sqrt2}\right)^2-2\cdot\frac{1}{\sqrt2}\cdot\left(1-\frac{1}{\sqrt2}\right)\cdot\frac{1}{\sqrt2}}\\&=\sqrt{\frac{1}{2}+\frac{3}{2}-\sqrt2-1+\frac{1}{\sqrt2}}=\sqrt{1-\frac{1}{\sqrt2}}.\end{align}$$