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Cauchy-Schwarz inequality
Task number: 4475
Use the Cauchy-Schwarz inequality to show that arbitrary real numbers a1,…,a5 satisfy:
4a1+3a2+6a3+4a4+2a5≤9√a21+⋯+a25
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}.