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

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.

• thanks (+1). have edited the question to make this clear – David Holden Jun 19 '17 at 1:56
• Looks like the original question has been resurrected. – dxiv Jun 19 '17 at 2:04
• hmm! not sure of the etiquette here. perhaps i should flag the resurrected post as a duplicate ;-) – David Holden Jun 19 '17 at 2:07
• Not sure, either. Looks like it must have been deleted and restored within $5$ minutes, since it doesn't even show in its edit history. @MlazhinkaShungGronzalezLeWy That question has 3 close votes FWIW it also has a valid answer posted by the OP, presumably after those close votes were casted. – dxiv Jun 19 '17 at 2:11
• @dxiv btw thanks for your sharp-eyed observation. i might not have realized what had happened until i was hauled before the inquisition – David Holden Jun 19 '17 at 2:24

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!

• bravo! a tour de force – David Holden Jun 19 '17 at 2:54
• @David Holden Your proof is beautiful! +1. – Michael Rozenberg Jun 19 '17 at 2:57

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 $y=\frac1x$ $$y^2f(x) = \left(x-\frac{a}2\right)^2 +\left(y-\frac{b}2\right)^2 + 2 - \frac{a^2}4 - \frac{b^2}4$$

• That is pretty neat! – TMM Jun 19 '17 at 2:30
• took a bit of head-banging to get there! – David Holden Jun 19 '17 at 2:32

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.

• The art of pretending to have a solution: Put "hint" and think hard on the text box. ha! – OR. Jun 19 '17 at 2:16
• @Miazhinka i admired the invocation of Cauchy-Schwartz here. that had not occurred to me, but it solves at a stroke the first vexing question, of how on earth to symmetrize the involvement of $a$ and $b$ – David Holden Jun 19 '17 at 2:28
• The Cauchy-Schwarz ansatz is great $\implies$ (+1) right now! May I propose a (SoS) shortcut, taking off in the second line? $$(1+x^2)^4\ge 8x^2(1+x^4)\;\iff\;(x^2+1)^4 - 8x^6-8x^2 = (x^2-1)^4\ge 0$$ This is a Sum of Squares certificate with only one summand $\ddot\smile$ – Hanno Nov 22 '17 at 12:41