<s>[INST] Non negative real values of $a,b,c$ The question is to find out the minimum value of non negative real values of $a,b,c$ given that the equation $$x^4+ax^3+bx^2+cx+1=0$$ has real roots.
I tried dividing both sides by $x^2$ the resultant equation becomes $$x^2+ax+b+\frac{c}{x} + \frac{1}{x^2}=0$$ I couldn't transfer this to a quadratic equation.If I take a=c then a quadratic equation may be formed by substituting $x+1/x=t$ with this assumption the equation reduces to $$t^2+at+b-2=0$$ Now we know that $t \geq 2$ when $x$ is positive and $t \leq -2$ when $x$ is negative. So we can check that if the roots of $t$ be in the interval $(-2,2)$ the corresponding value of $x$ will be imaginary.For this we have three conditions $f(-2)>0$,$f(2)>0$ and $\frac{-a}{2} \in (-2,2)$ [This comes from the fact that the minimum must be in the interval of the roots]  and the discriminant must be greater than or equal to zero.Using these conditions I get that (4,6,4) are the values of (a,b,c) However I am not sure how to proceed if $a\neq c$ .Any ideas?
 [/INST] The idea followed here is, for $a \ne c$, to apply a shift to $x$ in order to transform the equation into a form for which the OP has already given a solution. From that, we can infer back to the values of $a,b,c$. 
Applying a shift  $x = y+q$ gives 
$$
1 + c q + b q^2 + a q^3 + q^4 + (c  + 2 b q  + 3 a q^2+ 4 q^3) y + (b + 3 a q + 6 q^2) y^2 + (a + 4 q )y^3 + y^4 = 0
$$
Now you can select the $q = q(a,b,c)$ which solves
$$
c  + 2 b q  + 3 a q^2+ 4 q^3 = a' r\\
a + 4 q = a'\\
1 + c q + b q^2 + a q^3 + q^4 = r^2
$$
(coming back to that later).
Further, identify $b' =  b + 3 a q + 6 q^2$. This gives 
$$
0 = r^2+ a' r y + b'y^2 + a'y^3 + y^4 = y^2 (r^2/y^2+ a'r/ y + b'  + a'y + y^2)
$$
Setting $z = y/\sqrt r$ gives 
$$
0 = 1/z^2+ (a'/\sqrt r)/ z + b'/r  + (a'/\sqrt r) z +  z^2
$$
Now we can apply the OP's solution which gives
$a'/\sqrt r = 4$ and $b'/r = 6$.  Plugging that back into the equations above gives 
$$
c  + 2 b q  + 3 a q^2+ 4 q^3 = 4  \sqrt{r^3}\\
a + 4 q = 4 \sqrt r\\
1 + c q + b q^2 + a q^3 + q^4 = r^2\\
b + 3 a q + 6 q^2 = 6 r
$$
[One can check that for $a=c$, indeed $r=1$ and $q=0$ solve this system.]
Now here again  fourth-order equations are present so it won't be easy. We need to identify $a,b,c$ from these four equations by eliminating $q$ and $r$. (is there a fifth equation hidden somewhere? This would generally be necessary for eliminating $r$ and $q$, however: wait for what follows.)
We get $r = (q + a/4)^2$   and hence $b/6 +  a q/2 + q^2 = q^2 + a q/2 + a^2/16$ or $ b = 3a^2 / 8$.
Putting the results for $r$ and $b$ into the first system equation gives 
$
c  + 6 a^2 q / 8   + 3 a q^2+ 4 q^3 = a^3/16 + (3 a^2 q)/4 + 3 a q^2 + 4 q^3
$, so $c =a^3/16 $
Further, $r^2 = (q + a/4)^4 = a^4/256 + (a^3 q)/16 + (3 a^2 q^2)/8 + a q^3 + q^4 = 1 + c q + b q^2 + a q^3 + q^4 $. Plugging in $b$ and $c$ we have  $a^4/256  = 1  $ which gives $a = 4$ no matter which shift $q$.
So collecting we have $(a,b,c) = (4,6,4)$ as the smallest values even if we started without the condition $a=c$.
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