Answer:
\[|\vec{a}+\vec{b}{{|}^{2}}-{{(|\vec{a}|+|\vec{b}|)}^{2}}\] \[=|\vec{a}{{|}^{2}}+|\vec{b}{{|}^{2}}+2|\vec{a}||\vec{b}|\cos \theta -|\vec{a}{{|}^{2}}-|\] \[\vec{b}{{|}^{2}}-2\] \[|\vec{a}||\vec{b}|\] \[=-2|\vec{a}\text{ }\!\!|\!\!\text{ }\!\!|\!\!\text{ }\vec{b}|(1-\cos \theta )\] \[=-2|\vec{a}\text{ }\!\!|\!\!\text{ }\!\!|\!\!\text{ \vec{b} }\!\!|\!\!\text{ si}{{\text{n}}^{2}}\frac{\theta }{2}=\] a negative quantity Hence \[|\vec{a}+\vec{b}|\le |\vec{a}|+||\vec{b}|\].
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