= Solution
{c}
The hypothesis implies $v_1,v_2\ne0$, since the <zero vector> is a <scalar multiple> of every <vector>. Set
$$
t=\frac{\langle v_1,v_2\rangle}{\|v_2\|^2},\qquad w=v_1-tv_2.
$$
Since $v_1$ is not a <scalar multiple> of $v_2$, $w\ne0$. <Positive definiteness> and <bilinearity> of the real <inner product> give
$$
0<\langle w,w\rangle
=\|v_1\|^2-2t\langle v_1,v_2\rangle+t^2\|v_2\|^2
=\|v_1\|^2-\frac{\langle v_1,v_2\rangle^2}{\|v_2\|^2}.
$$
Multiplying by $\|v_2\|^2>0$ and taking <square roots> proves the strict <Cauchy-Schwarz inequality>
$$
|\langle v_1,v_2\rangle|<\|v_1\|\,\|v_2\|.
$$
Since every <real number> $s$ satisfies $s\leq|s|$, this proves, in particular,
$$
\boxed{\langle v_1,v_2\rangle<\|v_1\|\,\|v_2\|.}
$$
The argument derives the required <inequality> directly from the <inner product> axioms.
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