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
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products: Exercise $35$