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(A) The inequality $|\vec{a} \cdot \vec{b}| \leq |\vec{a}| |\vec{b}|$ is known as the Cauchy-Schwarz inequality.Case $1$: If $\vec{a} = \vec{0}$ or $\

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(A) The inequality $|\vec{a} \cdot \vec{b}| \leq |\vec{a}| |\vec{b}|$ is known as the Cauchy-Schwarz inequality.Case $1$: If $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$,then $|\vec{a} \cdot \vec{b}| = 0$ and $|\vec{a}| |\vec{b}| = 0$. Thus,$0 \leq 0$ holds true.Case $2$: If $\vec{a} 
eq \vec{0}$ and $\vec{b} 
eq \vec{0}$,we know that $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 	heta$,where $	heta$ is the angle between the vectors.Taking the absolute value on both sides,we get $|\vec{a} \cdot \vec{b}| = |\vec{a}| |\vec{b}| |\cos 	heta|$.Since the range of the cosine function is $[-1, 1]$,we have $|\cos 	heta| \leq 1$.Multiplying by $|\vec{a}| |\vec{b}|$ (which is positive),we obtain $|\vec{a}| |\vec{b}| |\cos 	heta| \leq |\vec{a}| |\vec{b}|$.Therefore,$|\vec{a} \cdot \vec{b}| \leq |\vec{a}| |\vec{b}|$ is always true.

For any two vectors $\vec{a}$ and $\vec{b}$,we always have $|\vec{a} \cdot \vec{b}| \leq |\vec{a}| |\vec{b}|$ (Cauchy-Schwarz inequality). Is this statement true or false?

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