Solution of a quadratic equation with polynomial coefficients Let $a$, $b$ y $c$ be real and positive numbers. If the following quadratic equation in $x$:
\begin{equation*}
(a+b+c)x^2-2(ab+bc+ca)x+ab^2+bc^2+ca^2=0
\end{equation*}
has at least one real solution. Determine the value of $\dfrac{a+5b}{c}$
I tried to apply algebraic identities, but I don't go anywhere.
 A: HINT: The $-2$ coefficient is somewhat suggestive, and the equation is homogeneous - can you decompose it into more tractable pieces (a quadratic would suggest squares might be a target)?

 It reduces to $$a(x-b)^2+b(x-c)^2+c(x-a)^2=0 \text { with } a,b,c\gt 0$$ in which all the terms are non-negative, and this implies $a=b=c$ if all are to be zero.

A: If it is a fact that any set $\{a,b,c\}$ of real positive number for which the polynomial has a real solution gives the same value for $\dfrac{a+5b}{c}$, then, since $a=b=c=1$ yields a polynomial in $x$ whose only solution is $x=1$, then it must be the case that $\dfrac{a+5b}{c}=6$.
The same result will be had for $a=b=c=r>0$
A: The positive numbers $a, b, c$ are very restricted so that the given equation admits a real root. 
In effect, since the coefficient of $x ^ 2$ is positive, the minimum of the given quadratic function must be less than or equal to zero. If $f(x)$ is the function we have this minimum at $x=\dfrac{ab+bc+ac}{a+b+c}$ and we must have 
 $f\left(\dfrac{ab+bc+ac}{a+b+c}\right)\le0$.
Calculation gives the necessary condition $$a^3c+b^3a+c^3b\le abc^2+bca^2+acb^2$$
and it can be shown that for positive numbers this inequality is verified only when $ a = b = c $ (which is left as a secondary problem) whereby the real root is double ($ x = a $).
Thus the required value is equal to $\color{red}6$.
A: Hint: By Cauchy Schwarz inequality:
$$(a+b+c)(ab^2+bc^2+ca^2)\geqslant (ab+bc+ca)^2$$
with equality iff $a=b=c$.  Now what was that determinant condition for the quadratic in $x$? 
