Hardest Geometry Problem Known To Man

If

$\pi(a+b)(1+\dfrac{3h}{10+\sqrt{4-3h}})=2a\pi(1-\displaystyle \sum_{i=1}^{100}\dfrac{(2i)!^2}{(2^i \times i!)^4}\times\dfrac{e^{2i}}{2i-1})$ Correct to five decimal places

Where $h=\dfrac{(a-b)^2}{(a+b)^2}$ $e=\dfrac{\sqrt{a^2-b^2}}{a}$ $a$ and $b$ are sides of an ellipse Give your answer in the form $a+b$ Given $a,b\le100$ Give the maximum possible value.