$$ \int_{0}^{1} \frac{dx}{1+x^a} > \frac{a}{a+1} $$
Could someone please help me prove the above inequality? I tried using differentiation under the integral sign, but I'(a) is not easy to compute either.
Edit:
Not relevant to the question, but here's what I could do:
As 1 + xa > 1, I could prove that the integral is definitely less than 1 for all values of a.
P.S.
Also, I tried to use LaTeX to format the question according to site's norms, but failed miserably. I'm a beginner, so please excuse it. In fact, it'd be great if someone edited this question using appropriate LaTeX code. Thanks a lot!
For $a > 0$, the series expansion $$(1+x^a)^{-1} = \sum_{k=0}^\infty (-1)^k x^{ak}$$ is valid on $x \in [0,1)$. Consequently we may write $$\int_{x=0}^1 \frac{dx}{1+x^a} = \sum_{k=0}^\infty \frac{(-1)^k}{1+ak},$$ and although not absolutely convergent, it is convergent for every $a > 0$ (why?); thus it must also equal $$\sum_{k=0}^\infty \frac{a}{(1+a(2k))(1+a(2k+1))},$$ whose summands are strictly positive and therefore the integral's value is bounded below by the first term of the series, which is $a/(a+1)$ as claimed.