I have difficulty showing this inequality $n^n(n+2)^{n+2}>(n+1)^{2n+2}$ holds for all natural numbers $n$. I thought about trying to show that
$$n^n(n+2)^{n+2}-(n+1)^{2n+2}=\frac{n^n-\frac{(n+1)^{2n+2}}{(n+2)^{n+2}}}{(n+1)^{n+1}}>0$$
It looks complicated and I don't know how to proceed further. Are there any useful identities involving consecutive natural numbers?
consider function $f(x)= (1+\frac {1}{x})^x$, Now it can be deduced that $f(x)$ is an increasing function for $x>0$,It is also deduced here How to prove $(1+1/x)^x$ is increasing when $x>0$?
now using the property of $f(x)$ that it is increasing for $x>0$
it follows
$$(1+\frac {1}{n+1})^{n+1}>(1+\frac {1}{n})^{n}$$
as $L.H.S$ is already greater than $R.H.S$, Multiplying it by a term Greater than "1" will not change the inequality
Here Multiply $L.H.S$ by $$(1+\frac {1}{n+1})$$ The inequality still holds true
it follows
$$(\frac {n+2}{n+1})^{n+2}>(\frac {n+1}{n})^{n}$$
cross multiplying terms, As all terms are positive
$$(n^n)(n+2)^{n+2}>(n+1)^{2n+2}$$
Q.E.D
$f(x) = \ln(x^x) = x \ln x$ is strictly convex on $(0, \infty)$ because the second derivative $f''(x) = 1/x$ is strictly positive. It follows that $$ f(n+1) < \frac 12 \bigl( f(n) + f(n+2) \bigr) \, , $$ i.e. $$ \ln ((n+1)^{n+1}) < \frac 12 \bigl(\ln (n^n) + \ln((n+2)^{n+2})\bigr) \, . $$ Now multiplication with $2$ and exponentiation gives the desired inequality $$ (n+1)^{2(n+1)} < n^n \cdot (n+2)^{n+2} \, . $$ The inequality holds not only for integers $n$ but for all positive real numbers.
Apply $AM \ge GM$ to $n$ copies of $\frac1{n}$ and $n+2$ copies of $\frac{1}{n+2}$, you get $$\verb/AM/ = \frac{n}{2n+2}\left(\frac1n\right) + \frac{n+2}{2n+2}\left(\frac{1}{n+2}\right) = \frac{1}{n+1}\stackrel{*}{\ge} \verb/GM/ = \left(\frac{1}{n}\right)^{\frac{n}{2n+2}}\left(\frac{1}{n+2}\right)^{\frac{n+2}{2n+2}}$$ The inequality $(*)$is strict because the numbers are not all the same. Rearrange above will get the inequality you seek.
