Solution of a quadratic equation with polynomial coefficients

Question

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.

Answer 1

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.

Answer 2

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 $a=b=c=r>0$

Answer 3

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 $\color{red}6$.

Answer 4

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 $x$?