By $\gamma(t) := \space(a\cos(t), b\sin(t))^T$ with $a,b > 0$ and $0 \le t \le 2\pi$ we define an ellipse.
Now I want to prove that the arc length is bound by $a$ and $b$ so that the following inequality hold:
$$\pi(a+b) \le L \le 2\pi \sqrt{a^2+b^2}$$
I tried using the definition and got:
$$L := \int_0^{2\pi} \sqrt{a^2\sin^2(t)+b^2\cos^2(t)}dt$$
which is equal to:
$$a\int_0^{2\pi} \sqrt{1-\frac{a^2-b^2}{a^2}\cos^2(t)}dt$$
Since the integral can't be solved by integrating I tried generating the Taylor Series with $c := \frac{\sqrt{a^2+b^2}}{a}$:
$$1-\frac{1}{2}c^2\cos^2(t)-\frac{1}{8}c^4\cos^4(t)... $$
However I don't know how this will help me.. Has anyone an idea or another way to prove this inequality?
Kind Regards
EDIT:
Upper Bound:
$$L = \int_0^{2\pi} \sqrt{a^2\sin^2(t)+b^2\cos^2(t)}dt \space \le \int_0^{2\pi} \sqrt{a^2*1+b^2*1}dt = 2\pi \sqrt{a^2+b^2}$$
Lower Bound:
See Macavity for a full answer.
So effectively we need to show $$\pi (a+b) \leqslant L = 4\int_0^{\frac{\pi}2}\sqrt{a^2\sin^2t + b^2\cos^2t } \: dt \leqslant 2\pi \sqrt{a^2+b^2}$$
For the right inequality, note for $x, y \geqslant 0$, we have $\sqrt{x^2+y^2} \leqslant x+y$, so $\sqrt{a^2\sin^2t + b^2\cos^2t } \leqslant a \sin t + b \cos t$, which integrated in $[0, \frac{\pi}2]$ gives the tighter bound $L \leqslant 4(a+b)$. Using QM-AM, to conclude if needed.
For the left inequality, note from Cauchy-Schwarz we get: $$(a^2\sin^2t + b^2\cos^2t)(\sin^2t + \cos^2t) \geqslant (a\sin^2t + b \cos^2t)^2$$ $$\implies L = 4\int_0^{\frac{\pi}2}\sqrt{a^2\sin^2t + b^2\cos^2t } \geqslant 4\int_0^\frac{\pi}2 (a \sin^2t + b \cos^2t) dt = \pi(a+b) $$