Determine the minimum of $a^2 + b^2$ if $a,b\in\mathbb{R}$ are such that $x^4 + ax^3 + bx^2 + ax + 1 = 0$ has at least one real solution I just wanted the solution, a hint or a start to the following question.
Determine the minimum of $a^2 + b^2$ if $a$ and $b$ are real numbers for which
the equation
$$x^4 + ax^3 + bx^2 + ax + 1 = 0$$
has at least one real solution.
Thanks in advance.
 A: Hint: Your quartic is palindromic. Divide through by $x^2$. We get the equation 
$$x^2+ax+b+\frac{a}{x}+\frac{1}{x^2}=0.$$
Make the substitution $x+\frac{1}{x}=t$. Then $x^2+\frac{1}{x^2}=t^2-2$, and we arrive at the equation
$$t^2+at+b-2=0.$$
Imagine solving the quadratic. We will get two not necessarily real values of $t$. The original equation has at least one real solution if and only if $t$ is real and has absolute value $\ge 2$. This is because, for example, for $x$ positive, the minimum value of $x+\frac{1}{x}$ is $2$.
Added, almost to the answers: There are two cases to consider, $a\ge 0$ and $a\lt 0$. Look at the case $a\ge 0$. The roots of the quadratic in $t$ are 
$$\frac{-a\pm \sqrt{a^2-4(b-2)}}{2}.$$
The root of largest absolute value is given by choosin the $-$ in $\pm$. We want this root to be $\le -2$. After some manipulation, that gives the inequality 
$$\sqrt{a^2-4(b-2)}\ge 4-a.$$
Squaring both sides and simplifying, we get $2a\ge b+ 2$. The minimum value of $a^2+b^2$ is attained when $2a=b+2$. That will give you the answer that has $a\ge 0$. The answer for negative $a$ is obtained by a similar calculation. The relevant root is then $t=\frac{-a+\sqrt{a^2-4(b-2)}}{2}$. 
Remark: I feel somewhat guilty at bashing the problem with standard machinery. But routine calculation does work. There is undoubtedly a simple way. However, simple can take time.  
A: [The following solution relies on the ideas in the previous two answers.]
Let the roots of $x^4+ax^3+bx^2+ax+1=0$ be given by $c, \frac{1}{c}, d, \frac{1}{d}$ where $c\in\mathbb{R}$, and 
let $t=c+\frac{1}{c}$ and $l=d+\frac{1}{d}$.  
Then $|t|\ge2$ since $c+\frac{1}{c}\ge2$ if $c>0$ and $c+\frac{1}{c}\le2$ if $c<0$.
Since $x^4+ax^3+bx^2+ax+1=(x-c)(x-\frac{1}{c})(x-d)(x-\frac{1}{d})=(x^2-tx+1)(x^2-lx+1)$,
we have that $a=-(t+l)$ and $b=2+lt$;  so
$a^2+b^2=(t+l)^2+(2+lt)^2=t^2+2lt+l^2+4+4lt+l^2t^2=(l^2+1)t^2+6lt+l^2+4$.
We consider two cases:
$\mathbf{A)}$ If $d\in\mathbb{R}$, then $|l|\ge2$ (for the same reason that $|t|\ge2$); so
$\;\;\;\;a^2+b^2=(l^2+1)t^2+6lt+l^2+4\ge20+6lt+8\ge4$ $\;$(since $|lt|\le4\implies lt\ge-4$).
$\mathbf{B)}$ If $d\not\in\mathbb{R}$, then $\frac{1}{d}=\bar{d}\implies|d|=1$, so $d=e^{i\theta}=\cos\theta+i\sin\theta$ for some $\theta\in\mathbb{R}$.  Then
$\;\;\;\;l=d+\frac{1}{d}=d+\bar{d}=2\cos\theta$, so $|l|\le2$.  
With $l$ fixed, let $f(t)=a^2+b^2=(l^2+1)t^2+6lt+l^2+4$.
Since the vertex of the graph of $f$ is at  $t=\frac{-3l}{l^2+1}$,
and $-2<\frac{-3l}{l^2+1}<2\;\;\;\;$ (since $2l^2+3l+2>0$ and $2l^2-3l+2>0$),
the minimum value of f for $|t|\ge2$ will occur when $t=2$ or $t=-2$:
1) If $t=2$, $a^2+b^2=4(l^2+1)+12l+l^2+4=5l^2+12l+8$, and 
$\;\;\;g(l)=5l^2+12l+8$ has its minimum when $l=-\frac{6}{5}$, with $g(-\frac{6}{5})=\frac{4}{5}$. 
$\;\;\;\;$(Notice that $|-\frac{6}{5}|\le2$).
2) If $t=-2$, $a^2+b^2=4(l^2+1)-12l+l^2+4=5l^2-12l+8$, and 
$\;\;\;g(l)=5l^2-12l+8$ has its minimum when $l=\frac{6}{5}$, with $g(\frac{6}{5})=\frac{4}{5}$. 
$\;\;\;\;$(Notice that $|\frac{6}{5}|\le2$).
From parts A and B, we can conclude that the minimum value of $a^2+b^2$ is $\;\;4/5$.
A: As has already been observed, if $c$ is a root of your equation, so is $c^{-1}$. So suppose your equation has roots $c,d,c^{-1},d^{-1}$.
We have
$$a=-(c+c^{-1} + d + d^{-1})$$
and
$$b=cc^{-1}+ cd+cd^{-1} +c^{-1}d +c^{-1}d^{-1} + dd^{-1}$$
$$=2+(c+c^{-1})(d+d^{-1})$$
Now if $c\in \mathbb{R}$, we must also have $d+d^{-1}\in \mathbb{R}$. This means $d=i$ or $d\in \mathbb{R}$.
Thus if we write $z=c+c^{-1}$ and $w=d + d^{-1}$, we see that your problem is a question of minimising
$$f=(w+z)^2 + (2+wz)^2$$
subject to the constraint $|w|\geq 2$ and $|z|\geq 2$ or $z=0$.
If $z=0$ the $f=w^2+4\geq8$. Consider instead $|z|\geq 2$. In this case $|zw|\geq 4$ so $f\geq (2-zw)^2\geq 4$. Thus we see that the minimal value of $f$ is 4. This is obtained when $z=2,w=-2$. And the equation is
$$x^4-2x^2+1$$
