Prove that $p(x)=x^4-x+\frac{1}{2}$ has no real roots. 
What is the simplest way to prove that the polynomial
  $p(x)=x^4-x+\frac{1}{2}$ has no real roots?

I did with Sturm's theorem:
$p_0(x)=x^4-x+\frac{1}{2}$
$p_1(x)=4x^3-1$
$p_2(x)=\frac34x-1$
$p_3(x)=-\frac{229}{27}$
The signs for $-\infty$ are $+,-,-,-$ and for $\infty$ are $+,+,+,-$. In the end $1-1=0$ real roots.
Can it be done faster?

You can use AM-GM to show that there are no real roots.  If $x< 0$, then $$x^4-x+\frac12> x^4+\frac12 > \frac12>0.$$
If $x\ge0$, then
$$x^4-x+\frac12 =\left(x^4+\frac1{4}+\frac1{8}+\frac1{8}\right)-x\geq 4\cdot\sqrt[4]{x^4\cdot \frac14\cdot\frac18\cdot\frac18}-x=x-x=0,$$
but the equality case doesn't occur because $\frac14\ne\frac18$.
You can even find the minimum value of $x^4-x+\frac12$ using AM-GM.  Note that when $x\ge 0$,
\begin{align}x^4-x+\frac12 &=\left(x^4+3\cdot\frac1{\sqrt[3]{4^4}}\right)-x+\left(\frac12-\frac{3}{\sqrt[3]{2^8}}\right)
\\&\geq 4\cdot\sqrt[4]{x^4\cdot \left(\frac1{\sqrt[3]{4^4}}\right)^3}-x+\left(\frac12-\frac{3}{\sqrt[3]{2^8}}\right)
\\&=x-x+\left(\frac12-\frac{3}{\sqrt[3]{2^8}}\right)=\frac12-\frac{3}{\sqrt[3]{2^8}}.
\end{align}
Note that the minimum value $\frac12-\frac3{\sqrt[3]{2^8}}\approx 0.02753$ is achieved if and only if $x=\frac{1}{\sqrt[3]{4}}\approx 0.62996$.
