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 $S=0$

  • 2
    $\begingroup$ Hm, $a_i$ are positive? $\endgroup$ – Fedor Petrov Jun 23 '19 at 12:15
  • 3
    $\begingroup$ 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. $\endgroup$ – Gerry Myerson Jun 23 '19 at 12:42
  • $\begingroup$ My computer is breakdown. I wrote this question by my mobilephone. I have corrected. Thanks You all. $\endgroup$ – Đào Thanh Oai Jun 23 '19 at 13:47
  • $\begingroup$ @GerryMyerson After I corrected, Maybe is not trivially true. $\endgroup$ – 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.


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