Question How do we prove that : $$n^n(n＋2)^{n＋1}\geq(n＋1)^{2n＋1}$$
My attempt $\text{prove by induction}$
in order to n=1 : correct becuase $3^2 \geq 2^3$
suppose $\forall n \in \mathbb{N} \colon n^n(n＋2)^{n＋1}\geq(n＋1)^{2n＋1} $ and prove that :$(n+1)^{n+1}(n＋3)^{n＋2}\geq(n＋2)^{2n＋3}$
I appreciate your interest
Consider $$ {n^n (n+2)^{n+1} \over n^n (n+1)^{n+1}} \geq {(n+1)^{2n+1} \over n^n (n+1)^{n+1}}, $$ which is $$ \left( 1 + \frac{1}{n+1} \right)^{n+1} \geq \left( 1 + \frac{1}{n} \right)^{n}. $$
Put $$ a_m = \left( 1 + \frac{1}{m} \right)^m. $$ By the binomial theorem, $$ \begin{aligned} a_m = {} & 1 + \frac{m}{m} + \frac{m(m-1)}{2!\, m^2} + \frac{m(m-1)(m-2)}{3!\, m^3} + \dots + \frac{m(m-1)(m-2) \cdots (m-(m-1))}{m!\, m^m} \\ = {} & 1 + 1 + \frac{1}{2!} \left( 1 - \frac{1}{m} \right) + \frac{1}{3!} \left( 1 - \frac{1}{m} \right) \left( 1 - \frac{2}{m} \right) + \dots + \frac{1}{m!} \left( 1 - \frac{1}{m} \right) \left( 1 - \frac{2}{m} \right) \cdots \left( 1 - \frac{m-1}{m} \right), \end{aligned} $$ which is not greater than $$ 1 + 1 + \frac{1}{2!} \left( 1 - \frac{1}{m+1} \right) + \frac{1}{3!} \left( 1 - \frac{1}{m+1} \right) \left( 1 - \frac{2}{m+1} \right) + \dots + \frac{1}{m!} \left( 1 - \frac{1}{m+1} \right) \left( 1 - \frac{2}{m+1} \right) \cdots \left( 1 - \frac{m-1}{m+1} \right), $$ which is in turn not greater than
$$ 1 + 1 + \frac{1}{2!} \left( 1 - \frac{1}{m+1} \right) + \frac{1}{3!} \left( 1 - \frac{1}{m+1} \right) \left( 1 - \frac{2}{m+1} \right) + \dots + \frac{1}{m!} \left( 1 - \frac{1}{m+1} \right) \left( 1 - \frac{2}{m+1} \right) \cdots \left( 1 - \frac{m-1}{m+1} \right) + \frac{1}{(m+1)!} \left( 1 - \frac{1}{m+1} \right) \left( 1 - \frac{2}{m+1} \right) \cdots \left( 1 - \frac{m-1}{m+1} \right) \left( 1 - \frac{m}{m+1} \right), $$ which is just $a_{m+1}$.
Hint: Write $$n^n(n＋2)^{n＋1}\geq(n＋1)^{2n＋1}$$ as $$\bigg(\frac{n}{n+1}\bigg)^n\ge \bigg(\frac{n+1}{n＋2}\bigg)^{n＋1}$$ or $$ \bigg(1+\frac{1}{n}\bigg)^n\le \bigg(1+\frac{1}{n＋1}\bigg)^{n＋1}. $$ Let $$ f(x)=\bigg(1+\frac{1}{x}\bigg)^x. $$ Try to show $f(x)$ is increasing.
The inequality is equivalent to
$$ \left( 1 + \frac 1 {n+1} \right)^{n+1} \ge \left( 1 + \frac 1 n \right)^n $$
To prove this we can use the AM-GM inequality. We have
$$ \begin{aligned} \left( 1 + \frac 1 {n+1} \right)^{n+1} &= \left( \frac 1 n + \frac 1 n + \cdots + \frac 1 {n+1} \right)^{n+1} \\ &\ge (n+1)^{n+1} \cdot \frac 1 {n^n(n+1)} \\ &= \left( 1 + \frac 1 n \right)^n \end{aligned} $$
Take $\log$ on each side to obtain $$n\log n+(n+1)\log(n+2)\geq (2n+1)\log (n+1) \iff \frac{n\log n+(n+1)\log (n+2)}{n+(n+1)}\geq \log (n+1)$$ Then, by Jensen's inequality, we have $$\log(n+1)=\log\left(\frac{(2n+1)(n+1)}{2n+1}\right)<\log \left(\frac{n\times n+(n+1)\times (n+2)}{n+(n+1)}\right)\leq \frac{n\log (n)+(n+1)\log(n+2)}{n+(n+1)}$$ We have used Jensen's inequality in the second inequality with $\phi(x)=\log x$.
$$n^n(n＋2)^{n＋1}\geq(n＋1)^{2n＋1} = n^n(n＋2)^{n＋1}\geq(n＋1)^{n + n＋1} \iff n^n(n＋2)^{n＋1}\geq(n＋1)^{n＋1} \cdot (n＋1)^n \\\iff \left(\dfrac{n + 2}{n + 1}\right)^{n + 1} \ge \left(\dfrac{n + 1}{n}\right)^n \\\iff\left( 1 + \frac{1}{n+1} \right)^{n+1} \geq \left( 1 + \frac{1}{n} \right)^{n}$$
If we show that $\displaystyle\left(1 + \frac{1}{k} \right)^{k}$ is increasing for $k > 0$, then we can prove $\displaystyle\left( 1 + \frac{1}{n+1} \right)^{n+1} \geq \left( 1 + \frac{1}{n} \right)^{n}$.
It suffices that $\displaystyle\left(1 + \frac{1}{k} \right)^{k}$ is increasing $\iff\displaystyle k\ln\left(1 + \frac{1}{k} \right)$ is increasing. Let $\displaystyle f(k) = k\ln\left(1 + \frac{1}{k} \right)$.
$$f'(k) = \left(\ln\left(1 + \dfrac{1}{k}\right) - \dfrac{1}{k + 1}\right)\cdot\left(1 + \dfrac{1}{k}\right)^k \\\left(\ln\left(1 + \dfrac{1}{k}\right) - \dfrac{1}{k + 1}\right)' = -\dfrac{1}{k\cdot(k + 1)^2} < 0 \\ \quad \\ \lim\limits_{k \to \infty}{f'(k)} \to 0 \, , \space\text{but $f'(k)$ is decreasing. Therefore $f'(k) > 0$ and we are done.}$$
You made a main step of the proof!
After your work we need to prove $$\left(\frac{n^2+2n}{(n+1)^2}\right)^{n^2+2n}\geq\left(\frac{n+1}{n+2}\right)^{n+2}$$ or $$\left(\frac{n+2}{n+1}\right)^{n+2}\geq\left(\frac{(n+1)^2}{n^2+2n}\right)^{n^2+2n},$$ which is true because $$\left(\frac{n+2}{n+1}\right)^{n+2}=\left(1+\frac{1}{n+1}\right)^{n+2}>e>\left(1+\frac{1}{n^2+2n}\right)^{n^2+2n}=\left(\frac{(n+1)^2}{n^2+2n}\right)^{n^2+2n}.$$
