## Calculus 8th Edition

(a) See the attached figure for a geometric interpretation. (b) $|a + b|\leq |a|+ |b|$
(a) As we are given: $|a + b|\leq |a|+ |b|$ Therefore, the given inequality tells that the length of one side of a triangle is less than the sum of the lengths of the other two sides. See the attached figure for a geometric interpretation. (b) By contradiction let us assume that $|a + b| \gt |a|+ |b|$ $|a + b| ^2\gt( |a|+ |b|)^2$ $(a+b) \cdot (a+b)= a .a+a.b+b.a+b.b=|a|^2+2(a \cdot b)+|b|^2$ $a \cdot b \gt |a||b|$ If $a \cdot b \gt 0$ then $|a \cdot b| a \cdot b$, so $|a \cdot b|\gt |a| |b|$ This contradicts the Cauchy-Schwarz Inequality. Therefore, the contradiction has been proved wrong . Hence, $|a + b|\leq |a|+ |b|$