Last updated at Dec. 8, 2016 by Teachoo

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Example 20 For any two vectors πο·― and πο·― , we always have | πο·― + πο·―| β€ | πο·―| + | πο·―| (triangle inequality). To Prove: | πο·― + πο·―| β€ | πο·―| + | πο·―| We first prove trivially, Therefore, the inequality | πο·― + πο·―| β€ | πο·―| + | πο·―| is satisfied trivially. Now, let us assume πο·― β 0ο·― & πο·― β 0ο·― | πο·― + πο·―|2 = ( πο·― + πο·―) . ( πο·― + πο·―) = πο·― . πο·― + πο·― . πο·― + πο·― . πο·― + πο·―. πο·― = πο·― . πο·― + πο·― . πο·― + πο·― . πο·― + πο·―. πο·― = πο·― . πο·― + 2 πο·―. πο·― + πο·―. πο·― = | πο·―|2 + 2 πο·―. πο·― + | πο·―|2 = | πο·―|2 + 2| πο·―|| πο·―| cos ΞΈ + | πο·―|2 β΄ | πο·― + πο·―|2 = | πο·―|2 + 2| πο·―|| πο·―| cos ΞΈ + | πο·―|2 We know that cos ΞΈ β€ 1 Multiplying 2| πο·―|| πο·―| on both sides 2| πο·―|| πο·―| cos ΞΈ β€ 2| πο·―|| πο·―| Adding | πο·―|2 + | πο·―|2 on both sides, | πο·―|2 + | πο·―|2 + 2| πο·―|| πο·―| cos ΞΈ β€ | πο·―|2 + | πο·―|2 + 2 | πο·―| | πο·―| | πο·― + πο·―|2 β€ (| πο·―| + | πο·―|) 2 Taking square root both sides | πο·― + πο·―| β€ (| πο·―| + | πο·―|) Hence proved.

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.