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It's true that: $$ \dfrac{x+y}{\sqrt{x^2+y^2}} \leq \sqrt{2} $$

Or, in perhaps a more useful form, for a vector $\mathbf{x} \in \mathbb{R}^2$, $$ x_1+x_2 \leq \sqrt{2} \cdot \lVert x \rVert $$

For $x,y>0$. This can be seen in many ways, perhaps the most obvious of which is to use polar coordinates, where it reduces to $\sin\theta + \cos\theta \leq \sqrt{2}$.

It's slightly non-trivial, but appears to be useful when you want to force the norm of a vector to show up.

Does this have a name? Does it generalize to arbitrary vector spaces or dimension?

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  • $\begingroup$ I don't know of a name, but you'll find the generalization to your inequality here. Indeed, it generalizes to any finite dimensional space. $\endgroup$ – Omnomnomnom Mar 25 '16 at 3:29
  • $\begingroup$ This is a relation between the taxicab norm and the Euclidean one, and as has been pointed out, generalizes to $R^n$ $\endgroup$ – Matematleta Mar 25 '16 at 3:44
  • $\begingroup$ It's called the extreme norm equivalence. I remember first seeing it in the work of an old lady mathematician Volzhvelskaya (who isn't very famous), but it's generalization, namely just norm equivalence on finite dimensional spaces, is credited to a whole lot of people. $\endgroup$ – астон вілла олоф мэллбэрг Mar 25 '16 at 4:11
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    $\begingroup$ Cauchy-Schwarz Inequality: $|x\cdot y|\leq \|x\|\cdot \|y\|$, for all $x,y$ in any number of dimensions. This is a special case when each co-ordinate of $y$ is $1$. $\endgroup$ – DanielWainfleet Mar 25 '16 at 5:10
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    $\begingroup$ One could say it's QM-AM inequality, which holds for any number of reals. It's easily put in the form:$$\frac{x+y}2\le \sqrt{\frac{x^2+y^2}2}$$ $\endgroup$ – Macavity Mar 25 '16 at 7:17

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