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


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 $p+1/p=-2$.



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 $(a,b)=\left(\pm\frac{4}{5},-\frac{2}{5}\right)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.