Sum with Maximum is Maximum of Sum

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Theorem

Let $a, b, c \in \R$ be real numbers.

Then:

$a + \max \set {b, c} = \max \set {a + b, a + c}$


Proof

Without loss of generality, there are two cases to consider:

$(1): \quad b \ge c$
$(2): \quad b < c$


First let $b \ge c$.

We have:

\(\ds b\) \(\ge\) \(\ds c\)
\(\ds \leadsto \ \ \) \(\ds a + b\) \(\ge\) \(\ds a + c\) Addition of Real Numbers is Compatible with Usual Ordering
\(\ds \leadsto \ \ \) \(\ds \max \set {a + b, a + c}\) \(=\) \(\ds a + b\) Definition of Maximum Element


Then:

\(\ds a + \max \set {b, c}\) \(=\) \(\ds a + b\) Definition of Maximum Element
\(\ds \) \(=\) \(\ds \max \set {a + b, a + c}\)

$\Box$


Now let $b < c$.

We have:

\(\ds b\) \(<\) \(\ds c\)
\(\ds \leadsto \ \ \) \(\ds a + b\) \(<\) \(\ds a + c\) Addition of Real Numbers is Compatible with Usual Ordering
\(\ds \leadsto \ \ \) \(\ds \max \set {a + b, a + c}\) \(=\) \(\ds a + c\) Definition of Maximum Element


Then:

\(\ds a + \max \set {b, c}\) \(=\) \(\ds a + c\) Definition of Maximum Element
\(\ds \) \(=\) \(\ds \max \set {a + b, a + c}\)

$\Box$


Thus the result holds in both cases.

Hence the result, by Proof by Cases.

$\blacksquare$


Sources