Let $ z_{1},z_{2},z_{3}$ be three complex numbers of moduli less than or equal to $ 1$. $ w_{1},w_{2}$ are two roots of the equation $$ (z-z_{1})(z-z_{2})+(z-z_{2})(z-z_{3})+(z-z_{3})(z-z_{1})=0$$ Prove that :for $ j =1,2,3$, $$\min\{|z_{j} - w_{1}|,|z_{j} - w_{2}|\}\leq 1$$ holds.
Following is geometry method
Proof: Let's denote $P(z)=(z-z_1)(z-z_2)(z-z_3)$. Obviously $w_1,w_2$ are the zeroes of $P'(z)$. It is well known (Gauss-Lucas) that for any polynomial $P$, the roots of $P'$ lie in the convex hull of the roots of $P$. Particularly, for a cubic polynomial we can go further; the Marden's theorem states that the roots of $P'$ lie in the foci of the Steiner Ellipse of the triangle formed by the roots of $P$, i.e. $z_1,z_2,z_3$. Thus, the problem becomes pure geometric. Given a triangle $ABC$, which lies inside of a circle with radius $1$, let $F_1,F_2$ be the foci of the Steiner ellipse. We must prove $\min (AF_1, AF_2)\le 1$. Let $G$ be the centroid of $ABC$, which is the midpoint of $F_1F_2$. It's enough to prove: $$3\min (AF_1, AF_2)^2 \le GA^2+GB^2+GC^2 \tag{1}$$ Indeed, since $GA^2+GB^2+GC^2 \le OA^2+OB^2+OC^2$ for any $O$, assuming $\min (AF_1, AF_2)> 1$ will lead us to a contradiction. Proving $\text{(1)}$ involves some calculations. One approach is to consider an equilateral triangle $A'B'C'$ with its incircle and to project it into $ABC$.
My Question:It is said this problem have algebra methods,can you help? Thanks