Minkowski's Inequality for Sums/Index 2

Theorem

Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \ge 0$ be non-negative real numbers.

Then:

$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^2}^{1 / 2} \le \paren {\sum_{k \mathop = 1}^n a_k^2}^{1 / 2} + \paren {\sum_{k \mathop = 1}^n b_k^2}^{1 / 2}$

Proof

\(\ds \sum_{k \mathop = 1}^n \paren {a_k + b_k}^2\)

\(=\)

\(\ds \sum_{k \mathop = 1}^n \paren {a_k^2 + 2 a_k b_k + b_k^2}\)

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 1}^n a_k^2 + 2 \sum_{k \mathop = 1}^n a_k b_k + \sum_{k \mathop = 1}^n b_k^2\)

\(\ds \)

\(\le\)

\(\ds \sum_{k \mathop = 1}^n a_k^2 + 2 \paren {\sum_{k \mathop = 1}^n a_k^2}^{1 / 2} \paren {\sum_{k \mathop = 1}^n b_k^2}^{1 / 2} + \sum_{k \mathop = 1}^n b_k^2\)

Cauchy's Inequality

\(\ds \)

\(=\)

\(\ds \paren {\paren {\sum_{k \mathop = 1}^n a_k^2}^{1 / 2} + \paren {\sum_{k \mathop = 1}^n b_k^2}^{1 / 2} }^2\)

The result follows from Order is Preserved on Positive Reals by Squaring.

$\blacksquare$

Also see

This result is a special case of Minkowski's Inequality for Sums, where $p = 2$.

Source of Name

This entry was named for Hermann Minkowski.

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: Exercise $\S 1.12 \ (6)$