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.

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 $\color{red}6$.
