Expression with $12$ degree polynomial always positive for real

Question

Proving $\displaystyle y^{12}-y^9+y^4-y+1>0$ forall real $y$

I am Trying to solve it using Discriminant Method

Like this solved by Lone Student

Prove that $x^8-x^5-x^4+x^2+x>-\frac{1}{3}$ for all real $x$

Let $\displaystyle P(y)=y^{12}-y^9+y^4-y+1$

I am Trying to convert it into Quadratic like polynomial with respect to $y$

$\displaystyle (y^8+1)y^4-(y^8+1)y+1$

But I did not understand How can I manipulate to form quadratic like polynomial in terms of $y$

Please have a look on that problem

Thanks

Answer 1

This is a solution using discriminant method.

To find polynomials $A,B,C,D$ such that $$P(y)=AD^2+BD+C$$ with $A\gt 0$ and $B^2-AC\lt 0$, I started with $D=y^m+n$ where $m,n$ are integers with $m\gt 0$.

Starting from integers $m,n$ where $m,|n|$ are small, I got $$P(y)=\underbrace{(y^6+y^3+1)}_{\text{positive}}(y^3-1)^2+(y+1)(y^3-1)+1$$ where $y^6+y^3+1=(y^3+\frac 12)^2+\frac 34\gt 0$.

Seeing this as a quadratic in $(y^3-1)$, the discriminant $D_1$ is given by $$D_1=(y+1)^2-4(y^6+y^3+1)$$

To prove that $D_1\lt 0$, it is sufficient to prove that $$Q(y):=4y^6+4y^3-y^2-2y+3\gt 0\tag2$$

To prove $(2)$, we can use discriminant method again.

We can write $$Q(y)=\underbrace{(y^2+1)}_{\text{positive}}(2y^2-1)^2+(2y+1)(2y^2-1)+3$$

Seeing this as a quadratic in $(2y^2-1)$, the discriminant $D_2$ is given by $$\begin{align}D_2&=(2y+1)^2-12(y^2+1) \\\\&=-8y^2+4y-11 \\\\&=-8\bigg(y-\frac 14\bigg)^2-\frac{21}{2}\lt 0\end{align}$$

Since $D_2\lt 0$, we can say that $(2)$ holds, so $D_1\lt 0$ follows.

Therefore, we can say that $P(y)\gt 0$ holds.

Answer 2

For all $y$, $P(y) = (y^8 + 1)(y^4 - y) + 1 = (y^8 + 1)(y^2 + y + 1)(y - 1)y + 1$. Clearly, $y^8 + 1 \geqslant 1 > 0$ and $y^2 + y + 1 = y^2 + y + \frac{1}{4} + \frac{3}{4} = \left(y + \frac{1}{2}\right)^2 + \frac{3}{4} \geqslant \frac{3}{4} > 0$ so $P(y) \geqslant 1 > 0$ when $(y - 1)y \geqslant 0$ i.e. when $y \leqslant 0$ or $y \geqslant 1$.

If $0 < y < 1$, as suggested @Macavity, $P(y) = 1 - y + y^4 - y^9 + y^{12} > 0$ because $1 > y$, $y^4 > y^9$ and $y^{12} > 0$.

In general, any real positive polynomial can be written as the sum of the squares of two real polynomials that don't vanish as in the article you shared (by induction on the degree thanks to a weird factorisation identity), but those polynomials are hard to find in general. To find them, you would need to be able to find the complex roots of $P$ (or at least the real decomposition in irreducible elements), which is pretty hard in general. I don't understand lone student's discriminant method (in particular, how he chooses his way to factorise).