<s>[INST] Proving that $\int_{0}^{1} \frac{dx}{1+x^a}>\frac{a}{1+a}, a>0$ Binding an integral from one or both sides is always interesting even if the integral is doable.
For instance, $$a>0, x \in (0,1) \implies  1>1-x^{2a} \implies \frac{1}{1+x^a} > 1-x^a.$$
Hence $$\int_{0}^{1} \frac{dx}{1+x^a}> \int_{0}^{1} (1-x^a)~ dx=\frac{a}{1+a}, a>0 ~~~(1)$$
The question is: How else (1) can be proved?
 [/INST] We can know that
$$
\begin{align*}
1-\int_0^1{\frac{1}{1+x^a}\text{d}x}&=\int_0^1{\left( 1-\frac{1}{1+x^a} \right) \text{d}x}
\\
&=\int_0^1{\frac{x^a}{1+x^a}\text{d}x}
\\
&<\int_0^1{x^a\text{d}x}
\\
&=\frac{1}{a+1}
\end{align*}
$$
Therefore $\displaystyle \int_0^1{\frac{1}{1+x^a}\text{d}x}>\frac{a}{a+1}$.
</s>