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

You can do better than the stated inequality starting from $$\log(1+x)<\frac{x(x+6)}{4x+6}$$ for all $x>0$. (Equality for $x=0$ and the difference between the right- and left hand side is strictly increasing as can be shown by taking derivatives.) With this we find $$ \frac{1}{2}\log\left(1+\frac{1}{x+\frac{1}{3}}\right)+x \log\left(1+\frac{1}{x}\right) < \frac38\left(\frac1{3x+1}+\frac1{x+1}\right)+\left(1-\frac3{6x+4}\right)=$$ $$1-\frac{3x}{4(3x+1)(x+1)(3x+2)}<1$$ for all $x>0$. In fact one can show in the same way that for any $\alpha > \tfrac16$ the inequality $$ \frac{1}{2}\log\left(1+\frac{1}{x+\alpha}\right)+x \log\left(1+\frac{1}{x}\right) < 1$$ holds for all $$x>\max\left(0,\frac{2-18\alpha^2}{18\alpha -3}\right).$$
