Proving roots are imaginary.

Question

If $a_1,a_2,\cdots,a_n\ (n\geq2)$ are real and $(n-1)a_1^2-2na_2<0$, then prove that at least two roots of the equation, $$x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_n=0$$ are imaginary.

Let $\{\alpha_i\}_{i=1}^n$ be the $n$ roots of the given equation. Then,$$\sum_i\alpha_i=-a_1;$$ also $$\sum_{i \ne j} \alpha_i\alpha_j = a_2.$$ It follows that \begin{align}(n-1)a_1^2-2na_2&= (n-1)\bigg(\sum_i \alpha_i\bigg)^2-2n\sum_{i \ne j} \alpha_i\alpha_j \\ &=n\bigg[{\bigg(\sum_i \alpha_i\bigg)^2-2\sum_{i \ne j} \alpha_i \alpha_j}\bigg]-\bigg(\sum_i \alpha_i\bigg)^2&\\&=n\sum_i \alpha_i^2-\bigg(\sum_i \alpha_i\bigg)^2 < 0.\end{align}

How to carry it further from here?

Answer 1

If $\alpha_i$ 's are all real then Cauchy - Schwarz inequality gives $(\sum \alpha_i)^{2} \leq n \sum (\alpha_i )^{2}$ which contradicts the inequality you have derived. Hence at least one root must be non-real. Since complex roots appear in conjugate pairs there must be two non-real roots. [Cauchy - Schwarz inequality gives $(\sum \alpha_i \beta_i )^{2} \leq \sum \alpha_i ^{2} \sum \beta_i ^{2}$. Put $\beta_i =1$ for all $i$.]

Answer 2

The claim is false. If the coefficients $f(x):=x^n+a_1x^{n-1}+\ldots +a_n$ match the inequality, then so do the coefficients of $g(x):=f(x+\epsilon)$ provided $|\epsilon|$ is small. But it cannot be the case that $g$ has imaginary roots for all small $\epsilon$ because that would mean $f$ has infinitely many roots.

Answer 3

$$n\sum \alpha_i^2=\begin{bmatrix}\alpha_1&\alpha_2&\cdots&\alpha_n\end{bmatrix}\begin{bmatrix}n&0&\cdots&0\\0&n&\cdots&0\\.\\.\\.\\0&0&\cdots&n\end{bmatrix}\begin{bmatrix}\alpha_1\\\alpha_2\\.\\.\\.\\\alpha_n\end{bmatrix}$$and $$(\alpha_1+\alpha_2+\cdots+\alpha_n)^2=\begin{bmatrix}\alpha_1&\alpha_2&\cdots&\alpha_n\end{bmatrix}\begin{bmatrix}1&1&\cdots&1\\1&1&\cdots&1\\.\\.\\.\\1&1&\cdots&1\end{bmatrix}\begin{bmatrix}\alpha_1\\\alpha_2\\.\\.\\.\\\alpha_n\end{bmatrix}$$if all the $a_i$s are real, the inequality imposes that $$\begin{bmatrix}n&0&\cdots&0\\0&n&\cdots&0\\.\\.\\.\\0&0&\cdots&n\end{bmatrix}-\begin{bmatrix}1&1&\cdots&1\\1&1&\cdots&1\\.\\.\\.\\1&1&\cdots&1\end{bmatrix}=\begin{bmatrix}n-1&-1&\cdots&-1\\-1&n-1&\cdots&-1\\.\\.\\.\\-1&-1&\cdots&n-1\end{bmatrix}$$must be negative definite while this is a contradiction since the set of eigenvalues is $$\{0,n\}$$ therefore at least two of $\alpha_i$'s are conjugate unreals since all the coefficients are real.

Answer 4

CS is a good way. Another way is to note that between every two real roots of a polynomial, its derivative needs to have a real root. Hence if $P(x) = x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_n$ has all real roots, then the quadratic $P^{(n-2)}(x)$ must have real roots. As $$P^{(n-2)}(x) = (n-2)!\binom{n}{n-2}x^2 + (n-2)!\binom{n-1}{n-2} a_1x+(n-2)!\binom{n-2}{n-2}a_2$$ the quadratic discriminant condition $\implies (n-1)a_1^2 - 2na_2\geqslant 0$, a contradiction.