Is $k(|a_1|+|a_2|+…+|a_n|) \le |b_1|+|b_2|+…+|b_n|+k|S|$ right?

Is the inequality as follows true?

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 $$S=0$$

• Hm, $a_i$ are positive? – Fedor Petrov Jun 23 '19 at 12:15
• You have specified $a_i>0$. So $|a_1|+\cdots+|a_n|=a_1+\cdots+a_n=S=|S|$, and your inequality is trivially true. – Gerry Myerson Jun 23 '19 at 12:42
• My computer is breakdown. I wrote this question by my mobilephone. I have corrected. Thanks You all. – Đào Thanh Oai Jun 23 '19 at 13:47
• @GerryMyerson After I corrected, Maybe is not trivially true. – Đào Thanh Oai Jun 23 '19 at 14:39

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.