<s>[INST] Inequality with integers I am trying to find an analytical way to prove that for integers $n\ge 6$,
$$\sqrt[n]{n+1}\le \frac{1}{n}+\frac{1}{\sqrt[n]{n+1}}+\frac12$$
 [/INST] Denote $x = \sqrt[n]{n+1}$. The desired inequality is written as
$\frac{-2nx^2 + (n+2)x + 2n}{2nx} \ge 0$.
It suffices to prove that $-2nx^2 + (n+2)x + 2n \ge 0$ which is written as
$$2n(x - x_1)(x - x_2)\le 0$$
where
$$x_1 = \frac{n+2 + \sqrt{17n^2 + 4n + 4}}{4n}, 
\quad x_2 = \frac{n+2 - \sqrt{17n^2 + 4n + 4}}{4n}.$$
Since $x > 0$ and $x_2 < 0$, it suffices to prove that $x \le x_1$, namely,
$$\sqrt[n]{n+1} \le \frac{n+2 + \sqrt{17n^2 + 4n + 4}}{4n}. \tag{1}$$
For $n = 6, 7, \cdots, 11$, (1) is verified directly.
For $n \ge 12$, since $\sqrt{17n^2 + 4n + 4} \ge 4n + 1$, it suffices to prove that
$$\sqrt[n]{n+1} \le \frac{n+2 + 4n + 1}{4n}$$
or
$$n+1 \le (1 + \tfrac{1}{4})^n (1 + \tfrac{3}{5n})^n.$$
By the binomial theorem, we have $(1 + \tfrac{1}{4})^n \ge 1 + \frac{1}{4}n + (\frac{1}{4})^2\frac{n(n-1)}{2}$
and
$(1 + \tfrac{3}{5n})^n \ge 1 + \frac{3}{5n}\cdot n = 1 + \tfrac{3}{5}$.
It suffices to prove that
$$n + 1 \le [1 + \tfrac{1}{4}n + (\tfrac{1}{4})^2\tfrac{n(n-1)}{2}] (1 + \tfrac{3}{5})$$
or
$$\frac{1}{20}(n-1)(n-12)\ge 0$$
which is true.
We are done.
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