<s>[INST] 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 $\displaystyle \int\limits_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$
 [/INST] Hint $:$ For maxima observe that $x \mapsto \frac {x} {x+1}$ is increasing on $[0,1].$ For minima observe that $x \mapsto \frac {1} {x+1}$ is decreasing on $[0,1].$
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