If $|a|\le1,|b|\le1,|c|\le1,|b-a|\le|c-a|$ then show that $\frac{|b-c||b-a|}{|2c-b-a|}\le1$ where $a,b,c \in \mathbb{C}$

Question

If $|a|\le1$,$|b|\le1$,$|c|\le1$ and $|b-a|\le|c-a|$ then show that $$\frac{|b-c||b-a|}{|2c-b-a|}\le1,$$ where $\{a,b,c\}\subset \mathbb{C}.$

This complex inequality is true, I am not able to prove it. I tried a lot of different approaches if there is any way you can think to solve this, it would help a lot.

Answer 1

Let $A(a)$, $B(b)$ and $C(c)$ in the Gauss plain.

Also, let $AB=z$, $AC=y$, $BC=x$, $M$ be a midpoint of $AB$

and $R$ be a radius of the circumcircle of $\Delta ABC$.

Thus, $z\leq y$, $R\leq1$ and we need to prove that $$BC\cdot AB\leq 2CM$$ or $$\sqrt{2x^2+2y^2-z^2}\geq xz$$ or $$2x^2+2y^2-z^2\geq x^2z^2.$$ Now, since $R=\frac{xyz}{4S_{\Delta ABC}}$, we obtain $$\frac{xyz}{4S_{\Delta ABC}}\leq1$$ or $$\sum_{cyc}(2x^2y^2-x^4)\geq x^2y^2z^2$$ or $$z^2(2x^2+2y^2-z^2)\geq(x^2-y^2)^2+x^2y^2z^2.$$ If $z=0$ then $a=b$, which gives that our inequality is true.

While for $z\neq0$ we obtain: $$2x^2+2y^2-z^2\geq\frac{(x^2-y^2)^2}{z^2}+x^2y^2\geq x^2z^2$$ and we are done!