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?

$f(x)=x^4$ is a convex function, hence its graph lies above the graph of the tangent line at $x=\frac{1}{2^{2/3}}$, whose equation is $g(x)=x-\frac{1}{2^{2/3}}+\frac{1}{2^{8/3}}$. $f(x)\geq g(x)$ implies 
$$ x^4-x+\frac{1}{2}\geq -\frac{1}{2^{2/3}}+\frac{1}{2^{8/3}}+\frac{1}{2}=\frac{4-3\sqrt[3]{2}}{8} $$
but $64>27\cdot 2$, so the RHS is positive and the LHS has no real zeroes.
