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I am having trouble showing the p-norm is in fact a norm on $\mathbb{R}^2$. We were first asked to show $f(x)=(1-x^p)^{1/p}$ is concave down on $[0,1]$. I now want to prove the triangle inequality for this norm, in other words: $$ (\lvert{x_1+y_1}\rvert^p+\lvert{x_2+y_2}\rvert^p)^{1/p} \leq (\lvert{x_1}\rvert^p+\lvert{x_2}\rvert^p)^{1/p}+(\lvert{y_1}\rvert^p+\lvert{y_2}\rvert^p)^{1/p} $$

I have done some reading on the subject and know I am being asked to prove Minkowski's inequality for $\mathbb{R}^2$. I am having trouble seeing where to start on this problem and I don't see where the fact $f(x)$ is concave down comes into the problem.

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Hint: We can show that if $f$ is "concave down", the line segment connecting the points $(x_0,y_0,f(x_0,y_0))$ and $(x_1,y_1,f(x_1,y_1))$ will lie beneath the graph given by $\{(x,y,f(x,y)):x,y \in \Bbb R\}$.

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  • $\begingroup$ To use this argument we would need to extend the definition of f to two dimensions. I think the natural way to do this would be to assume $x<y$ and let $f(x,y)=(1-(\frac{x}{y})^p)^{1/p}$ $\endgroup$ – mphy May 4 '15 at 18:38
  • $\begingroup$ Also, doing this leaves an expression that almost looks like what we want, but with a minus sign instead of a plus sign. $\endgroup$ – mphy May 4 '15 at 18:46

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