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I know the bound

$|a+b|^q \leq 2^{q-1} (|a|^q + |b|^q)$

See here for a proof. I can use this to show the bound

$|a+b+c|^q \leq 2^{q-1}(|a|^q + 2^{q-1}(|b| + |c|))$

and since the order is arbitrary I can do

$|a+b+c|^q \leq 2^{q-1}(\min(|a|^q + 2^{q-1}(|b| + |c|), |b|^q + 2^{q-1}(|a| + |c|), |c|^q + 2^{q-1}(|a| + |b|)))$

But is this the sharpest bound?

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  • $\begingroup$ You do mean $|a + b + c|^q$, not $|a + b + c|$, right? $\endgroup$ Jan 25, 2018 at 15:21
  • $\begingroup$ Yes, sorry. Updated. $\endgroup$
    – Tohiko
    Jan 25, 2018 at 15:25

1 Answer 1

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Setting $(a, b, c) = (1, 1, 2)$ gives $$|a + b + c|^q = 4^q$$ and $$2^{q-1} (|c|^q + 2^{q-1} (|a| + |b|)) = 2^{q-1} (2^q + 2^q) = 4^q$$ (the alternative bounds from rearranging $a, b, c$ are all larger), so the bound you give is optimal.

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