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Let $k > 0$, $a\_i$ Is a complex number for $1\le i\le n$ and let $$S:=a\_1+a\_2+....+a\_n$$ Suppose that $$b\_i:=S-ka\_i \quad\text{ for} \quad 1\le i\le n.$$ 
Then


$$k(|a\_1|+|a\_2|+...+|a\_n|) \le |b\_1|+|b\_2|+...+|b\_n|+k|S|$$


Equality if only if $S=0$



Is $k(|a_1|+|a_2|+...+|a_n|) \le |b_1|+|b_2|+...+|b_n|+k|S|$ right?
Now it is false: take $n=4$, $k=1$, $a\_1=a\_2=a\_3=1$, $a\_4=-2$. Then $S=1$, $b\_1=b\_2=b\_3=0$, $b\_4=3$ , RHS equals 4 and LHS equals 5.


