<s>[INST] simple proof that $\sqrt{1+\frac{1}{x+1/2}}(1+1/x)^x\le e$ It is well known that for $x>0$ that $\left(1+\frac{1}{x}\right)^x\le e\le\left(1+\frac{1}{x}\right)^{x+1}$ (see wikipedia). However, one can obtain the stronger inequality 
$$
\sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x\le e\le\sqrt{1+\frac{1}{x}}\left(1+\frac{1}{x}\right)^{x}
$$ 
The second inequality can be found in Proposition B.3 of "Randomized Algorithms", by Raghaven and Motwani (which itself refers to the book "Analytic Inequalities" by Mitrinović) , and can be proven straight-forwardly by calculus (showing a first derivative is non-negative and such).
While I can also prove the first inequality using familiar calculus methods, it is a bit messy (ultimately requiring that $\frac{1}{y+2}+\frac{1}{3y+2}\ge \frac{1}{y+1}$ for $y\ge 0$).
Does anyone know a "simple" proof of $\sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x\le e$? 
 [/INST] In response to @WimC, I'm posting my own proof of this inequality.  The first observation is to see that $\sqrt{1+\frac{1}{x+1/2}}(1+1/x)^x\le e$ for $x\ge 0$ is equivalent to (under $y:=1/x$), $$\left(1+\frac{2y}{y+2}\right)^{y/2}(1+y)\le e^y$$ for $y>0$.  One can then hope to establish this inequality for all $y \ge 0$.
Taking logarithms, this reduces to showing that $$f(y):=y-\ln(y+1)-\frac{y}{2}\ln(3y+2)+\frac{y}{2}\ln(y+2)$$ has $f(y)\ge 0$ for $y\ge 0$.  As $f(0)=0$, one needs only show the derivative $f'(y)\ge 0$ for $y\ge 0$.  Taking the derivative, one finds that
\begin{align*}
f'(y)&=1-\frac{1}{y+1}-\frac{1}{2}\ln\left(1+\frac{2y}{y+2}\right)+\frac{1}{2}\cdot\frac{y}{y+2}-\frac{1}{2}\cdot \frac{3y}{3y+2}\\
&\ge 1-\frac{1}{y+1}-\frac{1}{2}\cdot \frac{2y}{y+2}+\frac{1}{2}\cdot\frac{y}{y+2}-\frac{1}{2}\cdot \frac{3y}{3y+2}\\
&=\frac{y^2}{(y+1)(y+2)(3y+2)}
\end{align*}
where we have used $\ln(1+z)\le z$, then simplified.  This last expression is clearly non-negative, as desired.

My issue with the above proof is that the algebra is messy, and I don't gain any intuition on why the inequality is true.

In particular, I would ideally like to be able to get the best $\alpha$ so that
$$\sqrt{1+\frac{1}{x+1/2}}(1+1/x)^x\le e$$
is true for all $x>0$.  Using the above $y:=1/x$, this is equivalent to 
$$\sqrt{1+\frac{1}{1/y+\alpha}}(1+y)^{1/y}\le e$$
If we take $y\rightarrow \infty$ then as $1+y\sim e^{\ln y}$ we get $(1+y)^{1/y}\rightarrow 1$, so that we must have that for large $y$ that
$$\sqrt{1+\frac{1}{0+\alpha}}\le e$$
Rearranging, this says that $\alpha$ must obey $\alpha\ge\frac{1}{e^2-1}\approx\frac{1}{6.39}$ for large $y$.  To me it seems natural to expect that any such $\alpha$ works for all $y$, but that might be naive. 
</s>