Going off of this question: Maximum Proof (Average?) How can we prove that $max(|a|,|b|)\geq \frac{1}{2}(|a+b|)$?
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first try to prove $\max(|a|,|b|)\geq (|a|+|b|)/2$, I guess. – MAN-MADE Sep 13 '17 at 03:28
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@MANMAID I can prove that. How does this help? – user23899 Sep 13 '17 at 03:29
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How do the answers to the linked question fail to answer your question? In particular, this answer gives the entire argument! – Xander Henderson Sep 13 '17 at 03:30
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$|a+b|\leq |a|+|b|$ – MAN-MADE Sep 13 '17 at 03:30
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first for $|a|>|b|$ $$2max\{|a|,|b|\}=2|a|>|a|+|b| \to(\div 2) max\{|a|,|b|\}>\frac{|a|+|b|}{2}$$ second for $|a|<|b|$ $$2max\{|a|,|b|\}=2|b|>|b|+|a| \to(\div 2) max\{|a|,|b|\}>\frac{|a|+|b|}{2}$$ 3rd for $|a|=|b|$ $$2max\{|a|,|b|\}=2|a|=2|b|=|a|+|b| \to(\div 2) max\{|a|,|b|\}=\frac{|a|+|b|}{2}$$ as a result we have $$ max\{|a|,|b|\}\geq\frac{|a|+|b|}{2} $$ we know $$|a|+|b|\geq |a+b|$$ so $$max\{|a|,|b|\}\geq\frac{|a|+|b|}{2} \geq \frac{|a+b|}{2} $$
Khosrotash
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$(1/2)|x+y| \le (1/2)(|x| + |y|) \le$
$(1/2)( |x| + |y| +||x| - |y|| =$
$\max${$|x|,|y|$}.
Note :
$\max${$a,b$}$= (1/2)(a +b + |a-b|)$.
Peter Szilas
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