# If $\|\mathbf{OA}+k\mathbf{OB} \|=1$, prove that $\text{Area}(OACB) \leq \| \mathbf{OB} \|$

OACB is a parallelogram. In other words if $\left \|\mathbf{a}+k\mathbf{b} \right \|=1$ ($k\in\mathbb{R}$), prove that
$$\|\mathbf{a}\| \cdot \|\mathbf{b} \| \cdot \sin \theta \leq \|\mathbf{b} \|$$ where $\theta$ is the angle of the two vectors.

Any suggestion?

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If $\| a + kb \| = 1$ for all $k \in \mathbb{R}$, then in particular, you have $\| a + 0 b\| = \| a \| = 1$. The inequality follows immediately since $\sin x \leq |\sin x| \leq 1$.