Prove that $p(x)=x^4-x+\frac{1}{2}$ has no real roots.

Question

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?

Answer 1

Notice that

$$x^4-x+\frac{1}{2}=\left(x^2-\frac{1}{2}\right)^2+\left(x-\frac{1}{2}\right)^2 > 0$$

because both squares can not be zero at the same time.

Answer 2

The function is concave upward because its second derivative ($12x^2$) is positive.

The global minimum occurs when the derivative is zero: $4x^3-1=0$, $x=\sqrt[3]\frac14$.

The value there is $x^4-x+\frac12=\sqrt[3]\frac14\left(\frac14-1\right)+\frac12=\frac12-\frac34\sqrt[3]\frac14>0.$

[To see $\frac12>\frac34\sqrt[3]\frac14,$ cube both sides: $\frac18>\frac{27}{64}\frac14$ because $32>27$.]

Answer 3

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 $x=\frac{1}{\sqrt[3]{4}}\approx 0.62996$.

Answer 4

$f(x)=x^4$ 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.