if such that the polynomial $$P(x)=x^4+ax^3+2x^2+bx+1=0$$ has four real roots. prove or disprove $$a^2+b^2\ge 32?$$

I have solve this problem: if the polynomial $P(x)$ at least have one real root,then $a^2+b^2=8$,see a^2+b^2\ge 8,and when $P(x)=x^4+2x^3+2x^2+2x+1=(x+1)^2(x^2+1)$ but if the $P(x)$ have four real roots,I can't it.

  • $\begingroup$ I got that $a^2+b^2\ge 32/3.$ $\endgroup$ – Allawonder May 9 '20 at 17:29
  • $\begingroup$ Well, $a^2+b^2\geqslant 16$ is possible to show, but to get $32$ i need positivity restriction on the roots. $\endgroup$ – Macavity May 9 '20 at 17:36
  • $\begingroup$ Hello,what is example?@Macavity and @Allawonder, Thanks $\endgroup$ – geromty May 10 '20 at 0:34
  • $\begingroup$ Yeah, if the roots have an even number of signs, I'm good! $\endgroup$ – Allawonder May 10 '20 at 5:34
  • 1
    $\begingroup$ It's true that $a^2 + b^2 \geq 32$ holds. If $\Delta = \Delta_x P(x)$ is the discriminant of the polynomial, the curve $\Delta = 0$ separates the regions with different numbers of real roots: i.imgur.com/P6OVHmB.png. The singular points of the curve are $(0, 0)$ and $(\pm 4, \mp 4)$, the intersections with the circle are $(\pm 4, \mp 4)$. Therefore the picture correctly shows the regions where $P(x)$ has four real roots. $\endgroup$ – Maxim May 10 '20 at 12:45

Clearly, $x=0$ is not a root. WLOG, assume that $|b| \ge |a|$ (otherwise, letting $x = \frac{1}{y}$, we have $y^4 + by^3 + 2y^2 + ay + 1 = 0$; then swap $a$ and $b$).

Let $x_1 \le x_2 \le x_3 \le x_4$ be the four real roots. We split into two cases.

1) If $x_1x_4 > 0$, then either $x_i>0, \forall i$ or $x_i < 0, \forall i$. By AM-GM, we have $|a| = |x_1+x_2+x_3+x_4| \ge 4\sqrt[4]{x_1x_2x_3x_4} = 4$ and hence $a^2 + b^2 \ge 2a^2 \ge 32$.

2) If $x_1x_4 < 0$, then $x_1 \le x_2 < 0 < x_3 \le x_4$ since $x_1x_2x_3x_4 = 1$. From $x_1x_2 + x_2x_3 + x_3x_4 + x_4x_1 + x_1x_3 + x_2x_4 = 2$, we have \begin{align} x_1x_2 + x_3x_4 &= 2 - x_2x_3 - x_4x_1 - x_1x_3 - x_2x_4\\ &\ge 2 + 4\sqrt[4]{(-x_2x_3)\cdot (-x_4x_1) \cdot (-x_1x_3) \cdot (-x_2x_4)}\\ &\ge 6. \end{align} Then, we have \begin{align} a^2 &= (x_1 + x_2 + x_3 + x_4)^2 \\ &= x_1^2 + x_2^2 + x_3^2 + x_4^2 + 2 (x_1x_2 + x_2x_3 + x_3x_4 + x_4x_1 + x_1x_3 + x_2x_4)\\ &\ge 2x_1x_2 + 2x_3x_4 + 4\\ &\ge 16. \end{align} Thus, $a^2 + b^2 \ge 2a^2 \ge 32$.

We are done.


We will show that with some symmetry consideration we can either reduce to the case where all the roots have the same sign and then the result follows since if say $x_k >0$, $\sum {x_k} =|a| \ge 4(x_1..x_4)^{\frac{1}{4}} =4, a^2 \ge 16$ and then $\sum {1/x_k}=|b|$ and same applies so $b^2 \ge 16$ and we are done, or we must anyway have $|a|, |b| \ge 4$ and again the result follows.

Replacing $x \to 1/x$ changes $a$ into $b$ keeping the roots real so we can assume wlog $|a| \ge |b|$; replacing $x \to -x$ changes the signs of $a,b$ so we can assume wlog $a \ge 0$. But then if $x \ge 1, ax^3 \ge \pm bx$ since $a \ge |b|$ so $x^4+ax^3+2x^2+bx+1 >0$, while if $0 \le x \le 1, b=-c, a \ge c>0$ and we get the equation:

$x^4+ax^3+2x^2-cx+1=0, 0<c\le a$ and there is at least a root $0<x_0<1$ (sum of roots is negative so some are negative, product is $1$ so this means two are neagtive, two positive and our choices imply that the posiitive ones are in $(0,1)$)

Noting that the equation can be rewritten as:

$(x^2+2x-1)^2+(a-4)x^3+(4-c)x=0$ or $(x^2+2x-1)^2 =(c-4)x+(4-a)x^3$ we immediately get that if $c<4$ we must have $a<4$ too but then $4-a \le 4-c$ since $c \le a$ hence remembering that we have a root $0<x_0<1$ we get $0 \le (x_0^2+2x_0-1)^2 =(4-a)x_0^3-(4-c)x_0 < (4-a)x_0-(4-c)x_0 \le 0$ so a contradiction! Hence $|b|=c \ge 4, a \ge c \ge 4$ and we are done!


Following the steps given in Wikipedia, we can set the problem as
$$\text{Minimize}[a^2+b^2,\{a,b\}]$$ subject to the three inequality constraints $$P=16-3a^2 <0$$ $$D=-3 a^4+32 a^2-16 a b \leq 0$$ $$\Delta=-27 a^4-4 a^3 b^3+36 a^3 b-2 a^2 b^2+256 a^2+36 a b^3-512 a b-27 b^4+256 b^2 \leq 0$$ This is a quite simple task, leading to the solutions $a=\pm4$ and $b=\mp4$ and then $a^2+b^2 \geq 32$.

The symmetry, already underlined in other answers is obvious looking at the constraints.

All the above was done using calculus. It could be done without any if, as in many optimization problems of the kind, we assume that the solution will correspond to the saturation of inequality constraints. If this is the case, then $$D=0 \implies b=2 a\left(1-\frac{3 }{16}a^2\right)$$ $$\Delta=0 \implies a^2(a^2-16)^2(459 a^6-7776 a^4+43776 a^2-65536)=0$$ $a^2=0$ must be rejected since $P<0$ would not be satisfied.

Concerning the cubic in $a^2$, it shows only one real root which is $$a^2=\frac{32}{51} \left(9-\cosh \left(\frac{1}{3} \cosh ^{-1}(577)\right)\right) \approx 2.32647$$to be rejected since $P<0$ would not be satisfied $(2.32647 < \frac {16}3)$.

All of that makes that the only solution is $a^2=16$ to which corresponds $b^2=16$ too.

One thing I did not have time to do is to look at the shadow prices of this problem.


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