Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}} \geq \frac{n^2+3n}{2n+2}$ How to prove the following inequalities without using Bernoulli's inequality?

  
*
  
*$$\prod_{k=1}^{n}{\sqrt[k+1]{k}} \leq \frac{2^n}{n+1},$$
  
*$$\sum_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}} \geq \frac{n^2+3n}{2n+2}.$$
  

My proof:


*

*
\begin{align*}
\prod_{k=1}^{n}{\sqrt[k+1]{k}} &= \prod_{k=1}^{n}{\sqrt[k+1]{k\cdot 1 \cdot 1 \cdots 1}} \leq \prod^{n}_{k=1}{\frac{k+1+1+\cdots +1}{k+1}}\\
&=\prod^{n}_{k=1}{\frac{2k}{k+1}}=2^n \cdot \prod^{n}_{k=1}{\frac{k}{k+1}}=\frac{2^n}{n+1}.\end{align*}


*
$$\sum_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}} \geq n \cdot \sqrt[n]{\prod_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}}} \geq n \cdot \sqrt[n]{\frac{n+1}{2^n}}=\frac{n}{2}\cdot \sqrt[n]{n+1}.$$

It remains to prove that 

$$\frac{n}{2}\cdot \sqrt[n]{n+1} \geq \frac{n^2+3n}{2n+2}=\frac{n(n+3)}{2(n+1)},$$

or 

$$\sqrt[n]{n+1} \geq \frac{n+3}{n+1},$$
or 

$$(n+1) \cdot (n+1)^{\frac{1}{n}} \geq n+3.$$

We apply Bernoulli's Inequality and we have: 

$$(n+1)\cdot (1+n)^{\frac{1}{n}}\geq (n+1) \cdot \left(1+n\cdot \frac{1}{n}\right)=(n+1)\cdot 2 \geq n+3,$$
which is equivalent with:
$$2n+2 \geq n+3,$$ or

$$n\geq 1,$$ and this is true becaue $n \neq 0$, $n$ is a natural number. 



Can you give another solution without using Bernoulli's inequality?

Thanks :-)
 A: The inequality to be shown is 
$$(n+1)^{n+1}\geqslant(n+3)^n,
$$ 
for every positive integer $n$. Introduce the function $u$ defined by
$$
u(x)=(x+1)\log(x+1)-x\log(x+3),
$$
then standard computations yield 
$$
u'(x)=\frac3{3+x}+\log\left(\frac{1+x}{3+x}\right),\qquad
u''(x)=\frac{3-x}{(1+x)(3+x)^2},
$$
hence $u'$ increases on $(0,3)$ and decreases on $(3,+\infty)$. Since $u'(1)=\frac34-\log2\gt0$ and $u'(+\infty)=0$, $u$ is increasing on $(1,+\infty)$. Since $u(1)=0$, this yields $u(n)\geqslant0$ for every positive integer, QED.
Edit: Equivalently, one wants to prove that $(k+2)^{k-1}\leqslant k^k$ for every $k\geqslant2$, that is, $k+2\geqslant\left(1+\frac2k\right)^k$. If one knows that the RHS is increasing and converges to $\mathrm e^2\lt8$, this yields the result for every $k\geqslant6$. The cases $2\leqslant k\leqslant5$ can be checked manually.
A: The inequality to be shown is 
$$(n+1)^{n+1}\geqslant(n+3)^n,
$$ 
for every positive integer $n$. 
For $n = 1$ it is easy. For $n \ge 2$, apply AM-GM inequality to $(n-2)$-many $(n+3)$, 2 $\frac{n+3}{2}$, and $4$, we get
$$(n+3)^n < \left(\frac{(n-2)(n+3) + \frac{n+3}{2} + \frac{n+3}{2} + 4}{n+1}\right)^{n+1} = \left(n+1\right)^{n+1}$$
