Help me prove this inequality : How would I go about proving this?

$$ \displaystyle\sum_{r=1}^{n} \left( 1 + \dfrac{1}{2r} \right)^{2r} \leq n \displaystyle\sum_{r=0}^{n+1} \displaystyle\binom{n+1}{r} \left( \dfrac{1}{n+1} \right)^{r}$$

Thank you! I've tried so many things. I've tried finding a series I could compare one of the series to but nada, I tried to change the LHS to a geometric series but that didn't work out, please could someone give me a little hint? Thank you!

Here is the proof that  $(1+1/x)^x$ is concave for $x\ge 1$.
The second derivative of $(1+1/x)^x$ is $(1+1/x)^x$ times $$p(x)=\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)^2-\frac{1}{x(1+x)^2}$$
Now for $x\ge 1$, we have $$\ln(1+1/x)-\frac{2}{1+x}\le \frac{1}{x}-\frac{2}{1+x}=\frac{1-x}{x(1+x)}\le 0$$ and $$\ln(1+1/x)\ge \frac{1}{x}-\frac{1}{2x^2}\ge 0,$$ so \begin{align*}p(x)&= \ln^2(1+1/x)-\frac{2\ln(1+1/x)}{1+x}+\frac{1}{(1+x)^2}-\frac{1}{x(1+x)^2}\\
&=\ln(1+1/x)(\ln(1+1/x)-2/(1+x))+\frac{x-1}{x(1+x)^2}\\
&\le \left(\frac{1}{x}-\frac{1}{2x^2}\right)\left(\frac{1}{x}-\frac{2}{1+x}\right)+\frac{x-1}{x(1+x)^2}\\
&=-\frac{(x-1)^2}{2x^3(1+x)^2}\le 0
\end{align*}
proving that $(1+1/x)^x$ is concave for $x\ge 1$.
