<s>[INST] 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.
 [/INST] 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.  
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