My quadratics

Upon observation, taking real $a=b=c=K$ yields $3(x-K)^2 = 0$ , or $x = K$ a double real root. A more rigorous proof can be written as follows. Expanding out the above equation gives:

$3x^2 - 2(a+b+c)x + (ab + ac + bc) = 0$ (i)

which has roots:

$x = \frac{2(a+b+c) \pm \sqrt{4(a+b+c)^2 - 4(3)(ab+ac+bc)}}{6} = \frac{(a+b+c) \pm \sqrt{(a+b+c)^2 - 3(ab+ac+bc)}}{3}$ (ii).

If we desire real & equal roots, then the the discriminant in (ii) must equal zero, or:

$(a+b+c)^2 - 3(ab+ac+bc) = a^2 = b^2 +c^2 + 2(ab+ac+bc) - 3(ab+ac+bc) = 0 \Rightarrow a^2 + b^2 + c^2 - (ab+ac+bc) = 0$ (iii).

Taking the roots of (iii) (picking any of $a,b,c$ ) yields:

$a = \frac{(b+c) \pm \sqrt{(b+c)^2 - 4(1)(b^2 + c^2 - bc)}}{2} = \frac{(b+c) \pm \sqrt{-3b^2 + 6bc - 3c^2 }}{2} = \frac{(b+c) \pm \sqrt{-3(b-c)^2 }}{2}$ (iv).

We want $a,b,c \in \mathbb{R}$ , which occurs in (iv) iff $b = c = K$ to produce:

$a = \frac{(K+K) \pm \sqrt{-3(K-K)^2 }}{2} = \frac{2K}{2} = K$ (v).

Hence, $\boxed{a = b = c}$ is the required answer.

$\mathbb{Q}. \mathbb{E}. \mathbb{D}.$