The link here: $\max(a,b)=\frac{a+b+|a-b|}{2}$ generalization shows that $\max(a,b)=\frac{a+b+|a-b|}{2}$.
Is it true that $\max(|a|,|b|)=\frac{1}{2}(|a|+|b|)$?
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Michael Rozenberg
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user23899
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$\max(|a|,|b|)=\frac12(|a|+|a|)$ or $\max(|a|,|b|)=\frac12(|b|+|b|)$ and one can conclude that $\max(|a|,|b|)=\frac12(|a|+|b|)$ is between those two value unless $|a|=|b|$ in which case all 3 are equal. – kingW3 Sep 13 '17 at 14:58
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$\max\{|a|,|b|\}=\frac{1}{2}(|a|+|b|)$ is wrong. Try $a=1$, $b=3$.
By the way, the following statement is true. $$\max\{|a|,|b|\}\geq\frac{1}{2}(|a|+|b|).$$
Indeed, let $|a|\geq|b|$. We can do it because our inequality is symmetric.
Thus, $\max\{|a|,|b|\}=|a|$ and we need to prove that $$|a|\geq\frac{1}{2}(|a|+|b|)$$ or $$|a|\geq|b|,$$ which is our assuming.
Michael Rozenberg
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How is it true? I'm having a hard time seeing this without plugging in a bunch of numbers – user23899 Sep 13 '17 at 02:40
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Can you add details on this as well: $\max{|a|,|b|}\geq\frac{1}{2}(|a+b|)$. How does this differ? – user23899 Sep 13 '17 at 02:45
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As we already know, $$\max(a,b)=\frac{a+b+|a-b|}{2}.$$ Therefore $$\max(|a|,|b|)=\frac{|a|+|b|+||a|-|b||}{2}.$$
So if $\max(|a|,|b|)=\frac12(|a|+|b|),$ then \begin{align} \tfrac12(|a|+|b|) &= \frac{|a|+|b|+||a|-|b||}{2} \\ &= \tfrac12(|a|+|b|) + \tfrac12(||a|-|b||). \end{align}
Subtracting $\frac12(|a|+|b|)$ from both sides, $$ 0 = \tfrac12(||a|-|b||), $$ which implies that $|a| = |b|.$ In other words, the only way your formula can work is when you could pick either $|a|$ or $|b|$ and get the maximum value either way.
David K
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