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I want an inequality of the form : $\Vert a - b \Vert^2 \leq k.(\Vert a\Vert^2 + \Vert b\Vert^2)$ ? where k is a constant.

The norm in consideration is the euclidean norm, and $a$ and $b$ are vectors in $\mathbb{R} ^p$.

As a few people have replied below, its pretty straightforward with k = 2. But I was wondering if there is something tighter than that? Clearly k = 1 if a and b are independent (if random) or orthogonal.

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This is wrong for any norm: try $b=-a\ne0$. –  Did Jun 9 '12 at 14:08
    
Maybe you want $\Vert a - b \Vert^2 \leq 2(\Vert a\Vert^2 + \Vert b\Vert^2)$? –  Byron Schmuland Jun 9 '12 at 14:18
    
Or you want $\|a - b\| \leq \|a\| + \|b\|$? Or, taking squares, $\|a - b\|^2 \leq \|a\|^2 + \|b\|^2 + 2\|a\|\cdot\|b\|$. –  TMM Jun 9 '12 at 14:33
    
Thanks for your replies. I realize that my initial question was incorrect, so I modified it. –  NSR Jun 14 '12 at 13:06

1 Answer 1

No. Let $a \neq 0$ and $b = -a$. The left hand side is $4\|a\|^2$ and the right hand side is $2\|a\|^2$.

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