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Question Number 156691 by mathocean1 last updated on 14/Oct/21

$$Showthat$$$${\left(\underset{k=1}{\sum ^n}{x}_k{y}_k\right)}^2⩽\left(\underset{k=1}{\sum ^n}{x}_k^2\right)\times \left(\underset{k=1}{\sum ^n}{y}_k^2\right)t$$

Answered by puissant last updated on 14/Oct/21

$$P\left(\lambda \right)=\underset{k=1}{\sum ^n}{(x_k+\lambda y_k)}^2$$$$=\underset{k=1}{\sum ^n}(x_k^2+2\lambda x_k{y}_k+{\lambda }^2{y}_k^2)$$$$\Rightarrow P\left(\lambda \right)={\lambda }^2\underset{k=1}{\sum ^n}{y}_k^2+2\lambda \underset{k=1}{\sum ^n}{x}_k{y}_k+\underset{k=1}{\sum ^n}{x}_k^2$$$$P\left(\lambda \right)⩾0\Rightarrow \mathrm{\Delta }⩽0..$$$$\mathrm{\Delta }=4{\left(\underset{k=1}{\sum ^n}{x}_k{y}_k\right)}^2-4\left(\underset{k=1}{\sum ^n}{x}_k^2\right)\left(\underset{k=1}{\sum ^n}{y}_k^2\right)⩽0$$$$\Rightarrow {\left(\underset{k=1}{\sum ^n}{x}_k{y}_k\right)}^2⩽\left(\underset{k=1}{\sum ^n}{x}_k^2\right)\left(\underset{k=1}{\sum ^n}{y}_k^2\right)$$$$(InequalityofCauchy-Schwartz)...$$$$$$$$.............\mathcal{L}epuissant............$$

Commented by mathocean1 last updated on 22/Oct/21

$$thankslepuissant.$$

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