# 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?

• If you're allowed calculus, finding the global min seems easier to me. – user113102 Mar 6 '20 at 16:50

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.

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 $$32>27$$.]

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

$$f(x)=x^4$$ 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.