If $x,y,z$ are complex numbers such that $ |x| = |y|= |z| =1$, show that $$\vert 1+x |+ |1+y|+ |1+z| + |x+y| + |y+z| + |z+x| \geq 4$$
This is a problem from Romanian G.M. nr $11/2016$ and I couldn't solve it. I managed just to show that it is greater then $3$:
Let $s=|1+x|+|1+y|+|1+z|+|x+y|+|y+z|+|z+x|$.
Using the triangle inequality we obtain that $s\geq|3+3(x+y+z)|$.
On the other hand, using that $|x+y|=|−x−y|$ we obtain that $s\geq|3−(x+y+z)|$, so $3s\geq|9−3(x+y+z)|$ and so $4s\geq12$ and thus $s\geq 3$
By the triangle inequality we obtain: $$\sum_{cyc}\left(|1+x|+|x+y|\right)=$$ $$=\sum_{cyc}\frac{|1+x|+|1+y|+|-x-y|}{2}+\frac{|x+y|+|y+z|+|-x-z|}{2}\geq$$ $$\geq\sum_{cyc}\frac{|1+x+1+y-x-y|}{2}+\frac{|x+y+y+z-x-z|}{2}=4.$$ Done!