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 $\sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x\le e$? 

This does not prove the inequality for every $x$, but only when $x$ is succiently large.
Taking the $\log$ of the inequality:
$$\frac{1}{2}\log \Big(\frac{1}{x+\frac{1}{2}}+1 \Big)+x \log \Big(\frac{1}{x}+1 \Big) -1<0.$$
Using series expansion for $x=\infty$:
\begin{align}
& \frac{1}{2x+1}-\frac{1}{(2x+1)^2} +x\Big(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}\Big)+-1+o(\frac{1}{x^2})= \\
&=\frac{-4x^2+5x+2}{6(2x+1)^2x^2}+o(\frac{1}{x^2})
\end{align}
Hence the last quantity is less than $0$ for $x$ sufficiently large, and this implies that your inequality is true when $x$ is big enaugh.
