this is a hard problem.does any body have a good solution for it?

any solution or idea,please.

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 inellipse 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: \[3\min (AF_1, AF_2)^2 \le GA^2+GB^2+GC^2 \,\,\,\,\,\,\, \text{(1)} \] 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 $ABC$.

Let me share a proof I got 10 years ago which doesn't rely on Marden's theorem. Let $\omega\neq 1$ be a cube root of unity, \[u_1=z_1+\omega z_2+\overline{\omega}z_3,u_2=z_1+\overline{\omega}z_2+\omega z_3,\]$v_1$ and $v_2$ be a square root of $\frac{1}{3}u_1$ and $\frac{1}{3}u_2$ respectively, then by assumption we have $|u_1|,|u_2|\leq 3$, $|v_1|,|v_2|\leq 1$ and \[z_1-w_{1,2}=\frac{1}{3}(u_1+u_2\pm\sqrt{u_1u_2})=v_1^2+v_2^2\pm v_1v_2.\]Now we prove for any $v_1,v_2$ in the unit disc \[\min\{|(v_1-\omega v_2)(v_1-\overline{\omega}v_2)|,|(v_1+\omega v_2)(v_1+\overline{\omega}v_2)|\}\leq 1.\]WLOG we may assume $v_2=1$, $\Re{v_1}\geq 0$ and we prove that in this case \[|(v_1+\omega)(v_1+\overline{\omega})|\leq 1.\]It suffices to check the inequality for $v_1$ on the imaginary axis and the minor arc between $-\omega$ and $-\overline{\omega}$. In these cases the area of the triangle with vertices $v_1,-\omega,-\overline{\omega}$ is no greater than $\frac{\sqrt{3}}{4}$ and the angle between $v_1+\omega$ and $v_1+\overline{\omega}$ is in $[\frac{\pi}{3},\frac{2\pi}{3}],$ from which the inequality follows.

any Pure algebra Solutions???

This problem is a special case of the so-called Sendov-Ilieff conjecture: Let $P(z)$ be a polynomial ($\deg P \geq 2$), all of whose roots lie in the disk $|z| \leq 1$. If $z_0$ is one of the roots of $P(z)$, then the disk $|z-z_0| \leq 1$ contains at least one root of $P'(z)$. With this in mind, here's a proof based on the proof given in Prasolov's book Polynomials (1999). We first need a lemma. Lemma. Let $P(z)$ be a polynomial with degree $n$, and suppose some complex number $z_0$ satisfies \[ |P''(z_0)| \geq (n-1)|P'(z_0)|. \]Then $P'(z)$ has a root in the disk $|z-z_0|\leq 1$. Proof. If all roots $w_1,\ldots, w_{n-1}$ lie outside the disk, then \[ \left|\frac{P''(z_0)}{P'(z_0)}\right| \leq \sum_{j=1}^{n-1}\frac{1}{|z_0 - w_j|} < n-1, \]contradiction. $\square$ With this in mind, write $P(z) = (z-z_1)Q(z)$, where $Q(z) = (z-z_2)(z-z_3)$. Then \[ \frac{P''(z_1)}{P'(z_1)} = 2\frac{Q'(z_1)}{Q(z_1)} = 2\left(\frac{1}{z_1-z_2} + \frac{1}{z_1-z_3}\right) = \frac{2(2z_1- z_2 - z_3)}{(z_1 - z_2)(z_1 - z_3)}. \]Now consider the triangle $ABC$ with vertices $A = z_1$, $B = z_2$, and $C = z_3$. Then $|z_1 - z_2| = c$, $|z_1 - z_3| = b$, and $|2z_1 - z_2 - z_3| = 2m_a$, where $m_a$ is the length of the $A$-median. By the Lemma, the Sendov-Ilieff conjecture holds if $4m_a \geq 2bc$. Since the conjecture has the original assumption $R\leq 1$, we'll homogenize/strengthen and instead prove the inequality \[ 2Rm_a \geq bc\quad\text{ for all triangles }\quad\triangle ABC. \]But this is equivalent to \[ \frac 12 am_a \geq \frac{abc}{4R} = K = \frac12 ah_a, \]which follows since $m_a \geq h_a$. $\blacksquare$