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I cannot demonstrate this excersice about vectors: If $A$ and $B$ are mutually perpendicular vectors( different to zero vector) and "c" is any number show that :

$\|A +cB \| \geq \|A\|$.

Well I don´t know where to start, I was thinking about using the dot product but I don´t get how to use it, could anyone give me a solution or any idea?

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  • $\begingroup$ You know that $\|V\| = V \cdot V$ where $V\cdot V$ is the dot product of $V$ with itself, right? DId you try using that rule to rewrite both sides of the equation? $\endgroup$ – David K Sep 15 '16 at 20:03
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You need only note that \begin{align*} (A+cB, A+cB) &= (A, A) + c^2(B, B) + 2c^2(A, B)\\ &= (A, A) + c^2(B, B)\\ &\ge (A, A). \end{align*}

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Hint

$$\left\|A+cB\right\|^2=\left\|A\right\|^2+|c|^2\left\|B\right\|^2$$

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    $\begingroup$ Don't forget about the factor of $c$. $\endgroup$ – David K Sep 15 '16 at 20:00
  • $\begingroup$ @DavidK Thanks, good catch. $\endgroup$ – DonAntonio Sep 15 '16 at 20:01

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