Sum of r Powers is not Greater than r times Power of Maximum

From ProofWiki
Jump to navigation Jump to search

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:

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


Proof

By definition of the $\max$ operation:

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

Then:

$\forall i \in \set {1, 2, \ldots, r}: a_i \le a_k$

Hence:

\(\ds \forall i \in \set {1, 2, \ldots, r}: \, \) \(\ds a_i\) \(=\) \(\ds a_k\)
\(\ds \leadsto \ \ \) \(\ds \sum_{i \mathop = 1}^r a_i^n\) \(\le\) \(\ds \sum_{i \mathop = 1}^r a_k^n\)
\(\ds \) \(=\) \(\ds r a_k^n\)
\(\ds \) \(=\) \(\ds r a^n\) Definition of $a_k$

Hence the result.

$\blacksquare$


Sources