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I've got the number $\max\{a,b\}$ and $a,b\in \mathbb R$

$\max\{a,b\}=a, a\geq b$ or $\max\{a,b\}=b, a<b$
We can see that $\max\{a,b\}\leq c$ only if $a\leq c$ and $b\leq c$ Now, if $a,b,c,d \in \mathbb R$ , prove that $\max\{a+b,c+d\} \leq \max\{a,c\} + \max\{b,d\}$

I am really stuck here, I need this for my University.

Em.
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George K.
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  • I've got the number max{a,b} and a,b∈R

    max{a,b}=a, a≥b or max{a,b}=b, a<b We can see that max{a,b}≤c only if a≤c and b≤c Now, if a,b,c,d ∈ R , prove that max{a+b,c+d} ≤ max{a,c} + max{b,d}

    I am really stuck here, I need this for my University.

     – George K. Oct 19 '16 at 16:52
  • Formatting tips here. – Em. Oct 19 '16 at 16:59

2 Answers2

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Notice that $a\leq \max\{a,c\}$ and $b\leq \max\{b,d\}$ which means that $$a+b\leq \max\{a,c\}+\max\{b,d\}$$

Can you finish from here?

Ennar
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  • Can you please complete this proof? – agdhruv Feb 14 '18 at 20:21
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    @agdhruv, do the same for $c+d$. Then you have $a+b$ and $c+d$ are both bounded by $\max{a,c}+\max{b,d}$, so $\max{a+b,c+d}\leq \max{a,c}+\max{b,d}$. – Ennar Feb 15 '18 at 10:01
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An alternative is to consider $\max \{x,y\} = \frac{x + y + |x-y|}{2}$ and use the triangle inequality.

Dean C Wills
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