# Sum with Maximum is Maximum of Sum

## 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.

$\blacksquare$