0
$\begingroup$

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 $S=0$

$\endgroup$
4
  • 2
    $\begingroup$ Hm, $a_i$ are positive? $\endgroup$ Jun 23, 2019 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. $\endgroup$ Jun 23, 2019 at 12:42
  • $\begingroup$ My computer is breakdown. I wrote this question by my mobilephone. I have corrected. Thanks You all. $\endgroup$ Jun 23, 2019 at 13:47
  • $\begingroup$ @GerryMyerson After I corrected, Maybe is not trivially true. $\endgroup$ Jun 23, 2019 at 14:39

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.