$x^4 -ax^3 +2x^2 -bx +1$ has real root $\implies$ $a^2+b^2 \ge 8$

Question

it is requested to show that if the quartic polynomial $f(x) \in \mathbb{R}[x]$, defined by:

$$ f(x) = x^4 -ax^3 +2x^2 -bx +1, $$ has a real root, then $$ a^2 +b^2 \ge 8 $$ this question was asked by @medo, then deleted a few minutes ago. however having spent a little time on it, i think the problem seems sufficiently instructive to be worth resuscitating. it is not deep or difficult, but to find the right way of rewriting the polynomial to demonstrate the result is an interesting coffee-break challenge.

Answer 1

Let $x$ be a root. Thus, $x\neq0$ and $b=\frac{x^4-ax^3+2x^2+1}{x}$ and we need to prove that $$a^2+\frac{(x^4-ax^3+2x^2+1)^2}{x^2}\geq8$$ or $$(x^6+x^2)a^2-2(x^7+2x^5+x^3)a+x^8+4x^6+6x^4-4x^2+1\geq0,$$ for which it's enough to prove that $$(x^7+2x^5+x^3)^2-(x^6+x^2)(x^8+4x^6+6x^4-4x^2+1)\leq0$$ or $$(x^2-1)^4\geq0.$$ Done!

Answer 2

We have from Cauchy-Schwarz inequality: $((1+x^2)^2)^2=(x^4 + 2x^2 + 1)^2 = (ax^3+bx)^2 \le (a^2+b^2)(x^6+x^2) = (a^2+b^2)x^2(x^4+1)\implies a^2+b^2 \ge \dfrac{(1+x^2)^4}{x^2(x^4+1)}\ge 8 \iff (1+x^2)^4 \ge 8x^2(1+x^4) \iff (1+y)^4 \ge 8y(1+y^2), y = x^2 \ge 0$. Lastly, consider $f(y) = (1+y)^4 - 8y(1+y^2), y \ge 0\implies f'(y) = 4(1+y)^3 - 8-24y^2 = 4((1+y)^3 - 2 - 6y^2)= 4(y-1)^3$. Thus if $0 \le y \le 1 \implies f'(y) \le 0 \implies f(y) \ge f(1) = 0$. If $y \ge 1 \implies f'(y) \ge 0 \implies f(y) \ge f(1) = 0$. Either case $f(y) \ge 0\implies a^2+b^2 \ge 8$ as claimed.

Answer 3

Since the original question has now been resurrected, I may as well post my own thoughts on it. I tried a few things that didn't work, then hit upon the following approach, when $x \ne 0$ let $y=\frac1x$ $$ y^2f(x) = \left(x-\frac{a}2\right)^2 +\left(y-\frac{b}2\right)^2 + 2 - \frac{a^2}4 - \frac{b^2}4 $$