Proofs of pi

pi >= 2

$$\cos (x)-1=-{\int }_0^x\sin (t)dt$$ $$⟹2={\int }_0^{\pi }\sin (t)dt\le {\int }_0^{\pi }tdt=\frac{t^2}{2}{|}_0^{\pi }=\frac{{\pi }^2}{2}$$ $$⟹4\le {\pi }^2⟹2\le \pi$$

$\square$

pi >= 2\sqrt{2}

There is a simple way to do it with graph

So clearly, $2\pi \ge 4\sqrt{2}⟹\pi \ge 2\sqrt{2}$