Prove that $\left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1$ for all $N \geq 1$ I would like to prove that for all $N\geq 1$ we have,
$$\mathcal{P}(N) = \left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1.$$
Some basic simulations and worked out examples convince me that this inequality indeed holds true. I have tried to solve this problem by induction. Clearly, for $N=1$ we have,
$$\mathcal{P}(1) = \left(\frac{2}{3}\right)^{\tfrac{3}{4}}\cdot\left(\frac{2}{1}\right)^{\tfrac{1}{4}} \approx 0.8774 < 1.$$
Now assume the inequality holds for $N$, then for $N+1$ we have,
\begin{align}
\mathcal{P}(N+1) &=\left(\frac{2N+2}{2N+3}\right)^{\tfrac{2N+3}{2N+4}}\cdot\left(\frac{2}{1}\right)^{\tfrac{1}{2N+4}} \\
 &= \left(\left(\frac{2N}{2N+1}\right)\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)\right)^{\left(\frac{2N+1}{2N+2}\right)\left(\frac{2N+2}{2N+1}\cdot\frac{2N+3}{2N+4}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}\frac{2N+2}{2N+4}}\\[1em]
&= \left(\frac{2N+2}{2N+3}\right)^{\left(\frac{2N+1}{2N+2}\right)\left(1 + \frac{1}{2(N+1/2)(N+2)}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}\left(1 - \frac{1}{N+2}\right)}\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}\\
&= \small\left(\frac{2N+2}{2N+3}\right)^{\left(\frac{2N+1}{2N+2}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}}\cdot \left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\left(\frac{1}{2}\right)^{\frac{1}{N+2}} \left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}\\
&= \mathcal{P}(N) \cdot \left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\cdot\left(\frac{1}{2}\right)^{\frac{1}{N+2}} \cdot\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}
\end{align}
Now from here we know that the first three terms are all smaller than 1 ($\mathcal{P}(N) < 1$ by induction hypothesis). However the last term is larger than one. For the proof by induction to work out, we need that this last term cancels against,
$$\left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\cdot\left(\frac{1}{2}\right)^{\frac{1}{N+2}}.$$ But I do not see how it does. Any help is greatly appreciated.

If you could show that it is an increasing function of real positive $N$ and its limit is $1$ then that would be enough.
Showing the limit is $1$ as $N \to \infty$ looks easy enough: each component heads towards $1$
while the derivative seems to be $\mathcal P'(x)=\frac{{{\left( \frac{x}{2x+1}\right) }^{\frac{2x+1}{2x+2}}}\left( x\mathrm{log}\left( \frac{x}{2x+1}\right) +x+1\right) }{x{{\left( x+1\right) }^{2}}}$ which seems to be positive for positive $x$, so $\mathcal P(x)$ must be strictly less than $1$ for all positive real $x$ and thus $\mathcal P(N)$ must be strictly less than $1$ for all positive integer $N$
