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I found an inequality as following: Let $x, y, z$ be three complex numbers then:

\begin{equation*} \frac{1}{2}(|y+z-x|+|x+z-y| + |y+x-z|) \le |x| + |y|+|z|+\frac{1}{2}|x+y+z| \end{equation*} (1)

The inequality holds with equality if and only if $x+y+z=0$

Note that: I have a proof of the inequality (1).

My question: I am looking for a proof of conjecture as following:

Let $x, y, z$ in an inner product space $V$ then

\begin{equation*}\frac{1}{2}(\|y+z-x\|+\|x+z-y\| + \|y+x-z\|) \le \|x\| + \|y\|+\|z\|+\frac{1}{2}\|x+y+z\|\end{equation*}

where the norm ||z|| denotes the norm induced by the inner product

See also

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    $\begingroup$ So what if you take $x=1$ and $y=z=0$? If you put $3/2$ before the norms of single elements, on the right of the inequality, then it follows from Hlawka's inequality and the triangle inequality, after an appropriate change of variables. $\endgroup$
    – Fernando Muro
    Commented Jul 27, 2016 at 9:10
  • $\begingroup$ @FernandoMuro Thank to You , I am going to check and edit $\endgroup$
    – Oai Thanh Đào
    Commented Jul 27, 2016 at 9:32
  • $\begingroup$ @FernandoMuro, I corrected $\endgroup$
    – Oai Thanh Đào
    Commented Jul 27, 2016 at 9:59
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    $\begingroup$ I don't think this helps. You already claimed to have a proof for the previous obviously false formula. I think you should provide your temptative prove and move this question to MathSackExchange. $\endgroup$
    – Fernando Muro
    Commented Jul 27, 2016 at 10:06
  • $\begingroup$ Dear @FernandoMuro Please waiting me, I will post the proof in here. Because The original version is true. I don't want to post original formula, so I equivalent change (<=>) to new formula, but my change is false. But now the formula is correct because I checked my changes process $\endgroup$
    – Oai Thanh Đào
    Commented Jul 27, 2016 at 10:24

1 Answer 1

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It follows from the 1-dimensional case which you say is true: project everything to a randomly chosen line, apply 1d case and integrate.

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  • $\begingroup$ I will posted the proof in nex some days. Thank to You very much. $\endgroup$
    – Oai Thanh Đào
    Commented Jul 27, 2016 at 13:44
  • $\begingroup$ I don't know why the same topic is many vote up. But this is close topic. $\endgroup$
    – Oai Thanh Đào
    Commented Jul 27, 2016 at 14:08

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