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

I found a proof that seems well-motivated enough to satisfy me. The key lemma is the Hermite-Hadamard inequality (see wikipedia, or Old and new on the Hermite-Hadamard inequality).
Lemma (Hermite-Hadamard): Let $f:[a,b]\to\mathbb{R}$ be convex. Then $$ 
f\left(\frac{a+b}{2}\right)
\le\frac{1}{b-a}\int_a^b f(x)dx
\le \frac{f(a)+f(b)}{2}.
$$
This can be seen as an instantiation of the trapezoid rule for approximating integrals. The lower bound is Jensen's inequality, the upper bound follows from seeing (via convexity) that the curve $(x,f(x))$ is below the line $(a,f(a))\to(b,f(b))$ on the interval $[a,b]$. That Hermite-Hadamard is useful for the desired inequalities is in retrospect not surprising, as these inequalities can be used to prove Stirling's formula (up to the exact constant), and some proofs of Stirling's formula use Hermite-Hadamard/trapezoid-rule.
I'll first give a weaker result using Hermite-Hadamard which is pretty straightforward, then a proof of the exact result I wanted.

Applying Hermite-Hadamard to $\frac{1}{x}$ on $[x,x+1]$, and simplifying, we get that
$$
\frac{1}{x+\frac{1}{2}}
\le \ln\left(1+\frac{1}{x}\right)
\le \frac{1}{2}\left(\frac{1}{x}+\frac{1}{x+1}\right)
$$
Manipulating the lower bound, we see that
$$1\le \left(x+\frac{1}{2}\right)\ln\left(1+\frac{1}{x}\right)$$
which is equivalent to
$$e\le \sqrt{1+\frac{1}{x}}\left(1+\frac{1}{x}\right)^x.$$
Turning to the other direction, we want to show
$$
\sqrt{1+\frac{1}{x+1}}\left(1+\frac{1}{x}\right)^x\le e
$$
Taking logarithms,
$$
\frac{1}{2}\ln\left(1+\frac{1}{x+1}\right)+x\ln\left(1+\frac{1}{x}\right)\le 1
$$
Using that $\ln(1+y)\le y$, and the above upper bound via Hermite-Hadamard, we get
$$
\frac{1}{2}\ln\left(1+\frac{1}{x+1}\right)+x\ln\left(1+\frac{1}{x}\right)
\le \frac{1}{2}\cdot\frac{1}{x+1}+x\cdot \frac{1}{2}\left(\frac{1}{x}+\frac{1}{x+1}\right)
=\frac{1}{2}\left(\frac{x}{x}+\frac{x+1}{x+1}\right)=1
,$$
as desired.

To get the tighter bound, we can apply Hermite-Hadamard to $\ln(x)$ (which is concave, so the inequality is reversed) on $[x,x+1]$, so that (after integrating),
$$
\frac{1}{2}(\ln(x+1)+\ln(x))
\le \ln(x+1)+x\ln\left(1+\frac{1}{x}\right)-1
\le \ln\left(x+\frac{1}{2}\right)
$$
Manipulating the upper bound yields
$$
1
\ge x\ln\left(1+\frac{1}{x}\right)+\ln\left(\frac{x+1}{x+\frac{1}{2}}\right)
=x\ln\left(1+\frac{1}{x}\right)+\ln\left(1+\frac{1}{2}\cdot\frac{1}{x+\frac{1}{2}}\right).
$$
Exponentiating, this is
$$
e
\ge \left(1+\frac{1}{2}\cdot \frac{1}{x+\frac{1}{2}}\right)\left(1+\frac{1}{x}\right)^x
$$
Appealing to Bernoulli's inequality, $1+y/2\ge \sqrt{1+y}$
$$
e\ge \sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x,
$$
as desired. (One can also use the lower bound for Hermite-Hadamard on $\ln(x)$ to again get $e\le (1+1/x)^{x+1/2}$.)
