Cauchy-Schwarz inequality
Task number: 4475
Use the Cauchy-Schwarz inequality to show that arbitrary real numbers \(a_1,\dots,a_5\) satisfy:
\(4a_1 + 3a_2 + 6a_3 + 4a_4 + 2a_5 \le 9 \sqrt{a_1^2 + \cdots + a_5^2}\)
Solution
Choose \(\boldsymbol u=(a_1,\ldots,a_5)^{\mathrm T}\) and \(\boldsymbol v=(4{,}3,6{,}4,2)^{\mathrm T}\), then with respect to the standard inner product on \(\mathbb R^5\) we get \(\langle\boldsymbol u|\boldsymbol v\rangle=4a_1 + 3a_2 + 6a_3 + 4a_4 + 2a_5\), \(||\boldsymbol v||=\sqrt{4^2+3^2+6^2+4^2+2^2}=9\) and \(||\boldsymbol u||=\sqrt{a_1^2 + \cdots + a_5^2}\).
Then, according to the Cauchy-Schwarz inequality: \(4a_1 + 3a_2 + 6a_3 + 4a_4 + 2a_5 = \langle\boldsymbol u|\boldsymbol v\rangle \le |\langle\boldsymbol u|\boldsymbol v\rangle|\le ||\boldsymbol v||\cdot||\boldsymbol u||= 9 \sqrt{a_1^2 + \cdots + a_5^2}\).
