It may be the case that this is extremely trivial to prove, but I just cannot see it. I want to prove this:


for $0\leq a,\alpha,b,\beta\leq1$

I think it is also the case that

$|a\alpha-b\beta|\leq|a-b + \alpha-\beta|$

I actually THINK these might hold because I cannot provide any counter example.

Any kind of help would be extremely appreciated.

  • 1
    $\begingroup$ Also, the second inequality does not hold. Take strictly positive $a, \alpha$ such that $1\geq a+\alpha=b$ and $\beta=0$. You get $a\alpha\leq 0$. $\endgroup$
    – EuxhenH
    Mar 18, 2020 at 7:25

1 Answer 1


Apply the $\triangle$ inequality: $|a\alpha - b\beta| = |a\alpha - a\beta + a\beta - b\beta|\le |a||\alpha - \beta|+|\beta||a-b|\le |a-b|+|\alpha-\beta|$

  • $\begingroup$ I'm an idiot :)) thanks a lot $\endgroup$
    – Feri
    Mar 18, 2020 at 7:18
  • $\begingroup$ You are fine, no worry.... $\endgroup$
    – DeepSea
    Mar 18, 2020 at 7:19

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