Prove that $\frac{1}{2(n+2)}<\int_0^1\frac{x^{n+1}}{x+1}dx$ $\displaystyle\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\left[\frac{x^{n+2}}{(n+2)(x+1)}\right]_0^1+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$
$\displaystyle\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\frac{1}{2(n+2)}+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$
If we can prove that $\displaystyle\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$ is always greater than $0$ we can find the minima of the function. How can we prove that the term is always positive?
And how can we prove that $\displaystyle \int\limits_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$

If the function you are integrating is always positive, then the integral must be positive as well. Take a look at the function being integrated. We have
$$f(x) = \frac{x^{n+2}}{(x+1)^2(n+2)}$$ with $0\leq x \leq 1$. Given that $x\in [0,1]$, is it true that $f(x)$ is always positive?
I assume also that you mean to find the maxima of the $f(x)$ as above. To do this, we can rely on the fact that $f$ is a differentiable function. Hence we can use the first derivative test to find the local maxima, then compare the values of the local maxima to the endpoints $x=0$ and $x=1$.
