Inequality with integers I am trying to find an analytical way to prove that for integers $n\ge 6$,
$$\sqrt[n]{n+1}\le \frac{1}{n}+\frac{1}{\sqrt[n]{n+1}}+\frac12$$
 A: By the binomial expansion we obtain:
$$\left(1+\frac{\sqrt{2n-1}-1}{n-1}\right)^n\geq1+\frac{n(\sqrt{2n-1}-1}{n-1}+\frac{n(n-1)}{2}\cdot\left(\frac{\sqrt{2n-1}-1}{n-1}\right)^2=n+1.$$
Id est, $$\sqrt[n]{n+1}\leq1+\frac{\sqrt{2n-1}-1}{n-1}.$$
Thus, it's enough to prove that:
$$1+\frac{\sqrt{2n-1}-1}{n-1}\leq\frac{1}{n}+\frac{1}{1+\frac{\sqrt{2n-1}-1}{n-1}}+\frac{1}{2},$$
which is true for any $n\geq14$.
Can you end it now?
A: Here is an analytic proof using derivatives.
Let
$$f(x)=(x+1)^{1/x}-(x+1)^{-1/x}-{1\over x}=2\sinh\left({\ln(x+1)\over x}\right)-{1\over x}$$
It's enough to show that $f(6)\lt1/2$ and $f'(x)\lt0$ for $x\ge6$. The inequality $f(6)\lt1/2$ is not difficult to verify with a calculator. (It amounts to checking that $\sqrt[6]7\approx1.383$ is less than $(1+\sqrt{10})/3\approx1.387$.) As for the derivative, we have
$$\begin{align}
f'(x)&=2\left({1\over x(x+1)}-{\ln(x+1)\over x^2}\right)\cosh\left({\ln(x+1)\over x}\right)+{1\over x^2}\\
&={1\over x^2}\left(1-2\left(\ln(x+1)+{1\over x+1}-1\right)\cosh\left({\ln(x+1)\over x}\right) \right)\\
&\lt{1\over x^2}\left(1-2\left(\ln(x+1)+{1\over x+1}-1 \right)\right)\quad\text{if }x\ge2\\
&=-{1\over x^2}\left(2\ln(x+1)+{1\over x+1}-3 \right)
\end{align}$$
and it's easy to see that $g(x)=2\ln(x+1)+{1\over x+1}-3$ is an increasing function with $g(3)=2\ln4-{11\over4}\approx0.0226\gt0$, so $f'(x)\lt0$ for $x\ge3$. (It would have been enough to show $g(6)\gt0$, but it doesn't hurt to get a better bound on where the function starts decreasing; the actual maximum of $f$ occurs at around $2.005$.)
Remark: Computationally, the trickiest step here is showing $f(6)\lt1/2$; the inequality $g(3)\gt0$ amounts to showing $\ln2\gt11/16=0.6875$, and there are various ways to show that $\ln2\approx0.69$. To show $f(6)\lt1/2$ without simply letting a computer do all the work, let $u=\sqrt[6]7$, and note that
$$u+{1\over u}-{1\over6}\lt{1\over2}\iff3u^2-2u-3\lt0\iff{1-\sqrt{10}\over3}\lt u\lt{1+\sqrt{10}\over3}$$
so one need "merely" show that $7\cdot3^6\lt(1+\sqrt{10})^6$. Expanding the binomial and doing some arithmetic, this becomes $1226\lt403\sqrt{10}$, and this inequality is easily verified from $3.1^2=9.61\lt10$, so that
$$1226\lt1240=400\cdot3.1\lt403\sqrt{10}$$
A: 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.
