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 $\sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x\le e$? 
 A: 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.
A: 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 $$x>\max\left(0,\frac{2-18\alpha^2}{18\alpha -3}\right).$$
A: 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 $e\le (1+1/x)^{x+1/2}$.)
A: 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. 
