## 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: $$4a_1 + 3a_2 + 6a_3 + 4a_4 + 2a_5 = \langle\boldsymbol u|\boldsymbol v\rangle \le |\langle\boldsymbol u|\boldsymbol v\rangle|\le ||\boldsymbol v||\cdot||\boldsymbol u||= 9 \sqrt{a_1^2 + \cdots + a_5^2}$$.
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