Question: Prove that: $${p\over p+1}<\int\_{0}^{1}{1\over x^p+1}dx,\text{ where }p\in \mathbb{R}$$

**My attempt**:Write $${1\over x^p+1}=1-x^p+x^{2p}-x^{3}+...$$ $$\int\_{0}^{1}{1\over x^p+1}=\int\_{0}^{1}1-x^p+x^{2p}-x^{3}+...=(1-{1\over p+1})+({1\over 2p+1}-{1\over 3p+1})+....={p\over p+1}+{p\over (2p+1)(3p+1)}+...$$ Thus,$${p\over p+1}<\int\_{0}^{1}{1\over x^p+1}dx$$

Is this a correct proof?

Answer: 
The series expansion is unnecessary; it also leads to sticky questions about interchanging integration with the limit operation. (And the series expansion isn't valid for $p\le0$.) Better to chop the series after two terms, and establish the inequality:
$$
\frac1{1+x^p}\ge 1 - x^p
$$
for every $x$ between $0$ and $1$ and all $p$ real. Can you see why this is true?
For $p>-1$
$$ 1 = \int\_0^1 \frac{1+x^p}{1+x^p} \; dx < \int\_0^1 \frac{1}{1+x^p}\, dx + \int\_0^1 x^p \; dx = \int\_0^1 \frac{1}{1+x^p}\, dx + \frac{1}{p+1} $$ 
from which $$\int\_0^1 \frac{1}{1+x^p}\, dx > \frac{p}{p+1}.$$

For $p<0$, perhaps a more useful bound (?):
$$ \int\_0^1 \frac{1}{1+x^p} \; dx = \int\_0^1 \frac{x^{-p}}{x^{-p}+1}\, dx >\int\_0^1 \frac{x^{-p}}{2}\, dx = \frac{1}{2(1-p)}.$$