Power of Maximum is not Greater than Sum of Powers

Theorem

Let $a_1, a_2, \ldots, a_r$ be non-negative real numbers.

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Let $a = \max \set {a_1, a_2, \ldots, a_r}$.

Then:

$a^n \le a_1^n + a_2^n + \cdots + a_r^n$

Proof

By definition of the $\max$ operation:

$\exists k \in \set {1, 2, \ldots, r}: a_k = a$

Hence:

\(\ds a^n\)

\(=\)

\(\ds a_k^n\)

\(\ds \)

\(\le\)

\(\ds a_1^n + a_2^n + \cdots + a_k^n + \cdots + a_r^n\)

Hence the result.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $1$: Review of some real analysis: Exercise $1.5: 11$