Condition for quartic polynomial coefficients given at least one real root

Question

Find the minimum possible value of $a^2+b^2$ where $a$, $b$ are two real numbers such that the polynomial $$x^4+ax^3+bx^2+ax+1,$$ has at least one real root.

My attempt: Let p be a real root. Therefore $p^4 + a(p^3) + b(p^2) + ap + 1 = 0$.

Divide both sides by $p^2$, noting that $p$ cannot be zero since $P(0) = 1$, we get $$p^2 + ap + b + a/p + 1/(p^2) = 0.$$

I rearranged this into a quadratic; i.e. $$(p+1/p)^2 + a(p+1/p) + (b-2) = 0.$$

If $p$ is real, $p+1/p$ is real, so discriminant is non-negative. setting discriminant $\geq0$, I get:

$$a^2 - 4(b-2)\geq0,$$

$$a^2 \geq 4b-8,$$

$$a^2 + b^2 \geq b^2 + 4b - 8 = (b+2)^2 - 12 \geq -12$$ But this is useless because it is obvious $a^2 + b^2 \geq 0$.

I believe this is because I also need to use the fact that $|p+1/p| \geq 2$, however I am unsure how to use this inequality with the discriminant.

Thanks in advance

Answer 1

Replacing $a$ with $-a$ only changes the signs of the zeros, so w.l.o.g. we can assume that $a\ge0$. The quadratic with the unknown $p+1/p$ (nice trick, BTW!) that you derived gives $$ p+\frac1p=\frac{-a\pm\sqrt{a^2-4(b-2)}}2. $$ Given the assumption $a\ge0$ we see that of these two alternatives the solution with a minus sign gives the larger value to $|p+1/p|$. Therefore the condition $|p+1/p|\ge2$, now rewritten as $p+1/p\le-2$, gives us another constraint $$ -a-\sqrt{a^2-4(b-2)}\le-4. $$ This is equivalent to $$ 4-a\le\sqrt{a^2-4(b-2)}.\qquad(*) $$ We shall see that there are solutions with $a^2+b^2<16$, so w.l.o.g. we can further assume that $a<4$, so $4-a>0$ and we can square both sides of $(*)$ arriving at $$ 16-8a+a^2\le a^2-4b+8\implies a\ge (b+2)/2. $$ Again, later developments will reveal that $b+2$ must be positive at the sought minimium, so to minimize $a^2$ we must have equality here, i.e. $a=(b+2)/2$. Thus we really want to minimize $$ a^2+b^2=\frac14[(b+2)^2+4b^2]=\frac14[5b^2+4b+4]. $$ It is trivial to show that this has a minimum at $b=-2/5$. The corresponding $a=(b+2)/2=4/5$, and at this point we have $$ a^2+b^2=\frac{4^2+2^2}{25}=\frac45. $$

If either the assumption $4-a\ge0$ or the assumption $b+2\ge0$ (that I made while looking for this candidate point) were invalid, then clearly $a^2+b^2$ would have a larger value, so we can dismiss those possibilities.

As a final check we see that when $a=4/5, b=-2/5$ your polynomial has a double root at $x=-1$, not unexpectedly matching with $p+1/p=-2$.

Answer 2

Hint:

Put $t=x+\frac{1}{x}$ then:

$x^4+ax^3+bx^2+ax+1=0$ has a real root if and only if $t^2+at+(b-2)=0$ has a real root $t$ such that $|t| \ge 2$, which is equivalent to \begin{equation}\tag{*} \left[\begin{matrix} \frac{-a-\sqrt{a^2-4(b-2)}}{2} \le -2 \\ \frac{-a+\sqrt{a^2-4(b-2)}}{2} \ge 2 \end{matrix}\right. \end{equation}

Using this: $$A \le \sqrt{B} \Longleftrightarrow \left[\begin{matrix} \left\{\begin{matrix} B \ge 0 \\ A < 0 \end{matrix}\right. \\ \left\{\begin{matrix} A \ge 0 \\ A^2 \le B \end{matrix}\right. \end{matrix}\right. $$

After some operations we have $(*)$ is equivalent to

$$\left[\begin{matrix} (1)\left\{\begin{matrix} a \ge 4 \\ a^2-4(b-2) \ge 0 \end{matrix}\right. \\ (2)\left\{\begin{matrix} a \le -4 \\ a^2-4(b-2) \ge 0 \end{matrix}\right. \\ (3)\left\{\begin{matrix} a < 4 \\ 2a -b \ge 2 \end{matrix}\right. \\ (4)\left\{\begin{matrix} a > -4 \\ 2a+b \le -2 \end{matrix}\right. \end{matrix}\right.$$

For $(1)$ and $(2)$: $a^2+b^2 \ge 4^2+0 =16$.

For $(3)$: using $5(a^2+b^2) = (2a-b)^2+(a+2b)^2$ we have $a^2+b^2 \ge \frac{4}{5}$, attained when $a=\frac{4}{5},b=-\frac{2}{5}$.

For $(4)$: using $5(a^2+b^2) = (2a+b)^2+(a-2b)^2$ we have $a^2+b^2 \ge \frac{4}{5}$, attained when $a=-\frac{4}{5},b=-\frac{2}{5}$.

Conclusion: the minimum value of $a^2+b^2$ is $\frac{4}{5}$, attained when $(a,b)=\left(\pm\frac{4}{5},-\frac{2}{5}\right)$.