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    • Thread Starter

    Is it true that

    \sqrt{a^2 + b^2} \geq  |a| + |b|?

    How do you prove it?

    I think it's something used in solutions to a much longer question, but I'm not sure.
    • PS Helper

    PS Helper
    Square both sides and rearrange; if you notice that all the rearranging you do uses \Leftrightarrow rather than just \Rightarrow (since everything's nice and positive), then you're good to go.

    EDIT: Shouldn't that be \le? If you think of (a,b) as a point in \mathbb{R}^2 then what the inequality is (or should be) saying is that the distance from the origin to the point (a,b) is shorter than (or equal to) the horizontal distance plus the vertical distance.
    • Thread Starter

    Well, what I'm really trying to do is to understand this:

    x = (x1, x2) and a = (a1, a2)

    |x1||x2 - a2| + |x1 - a1||a2| <= (|x1| + |a2|)(||x-a||)

    I think what they've said is that |x1||x2 - a2| <= (|x1| + |a2|)|x2 - a2|

    And |x1 ? a1||a2| <= |x1 ? a1|(|a2| + |x1|)

    So |x1||x2 - a2| + |x1 - a1||a2| <= (|x2| + |a2|)(|x1 - a1| + |x2 - a2|)

    But where do you get that (|x1 - a1| + |x2 - a2|) <= ||x - a||?
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Updated: April 4, 2011

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