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Im taking mathematical analysis 1, and one of the problems asks

Prove that $$\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2} \leq \sqrt{a_1^2+a_2^2+a_3^2}+\sqrt{b_1^2+b_2^2+b_3^2}$$ where $a_j,b_j \in \mathbb{R}, j=1,2,3$

My attempt at a solution:

Each radicand is greater than or equal to zero, so I squared both sides and expanded and eliminated terms to get:

$$a_1b_1-a_2b_2-a_3b_3\geq-\sqrt{(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)}$$

I know that the right hand side is definitely a negative number (or 0 if all terms are 0) but I believe this path leads nowhere. I can't go about squaring again because it's unclear whether or not the left hand side is positive or negative.

Any advice would be tremendously appreciated.

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Consider three points in $\mathbb{R}^3$: $$ A=(a_1,a_2,a_3),\qquad B=(b_1,b_2,b_3),\qquad C=(0,0,0).$$ The given inequality can be read as $$ \|A-B\| \leq \|A-C\|+\|C-B\| $$ that is trivial.

Anyway, $$\left|a_1 b_1+a_2 b_2+b_3 c_3\right|\leq\sqrt{(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)}$$ is a consequence of the Cauchy-Schwarz inequality.

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  • $\begingroup$ I prefer $AB\leq OA+OB$. $\endgroup$
    – Takahiro Waki
    Sep 14, 2016 at 22:13
  • $\begingroup$ This could be a stupid question, but what really is meant by $A-B$, if $A$ and $B$ are points? The problem makes sense geometrically now that I think about it; it's really just an application of the triangle inequality. But I feel like a geometric argument isn't good for some reason $\endgroup$
    – user225028
    Sep 15, 2016 at 0:59
  • $\begingroup$ @johnmorrison: $A-B$ is the coordinate-wise difference of $A$ and $B$, hence $\|A-B\|$ is simply the euclidean distance between $A$ and $B$. $\endgroup$
    – Jack D'Aurizio
    Sep 15, 2016 at 1:50

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