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The statement I'm trying to prove is:

$(x+y)^c\le x^c+y^c$ whenever $0\le x,y$ and $0\le c\le1$.

This comes up in the proof that $|x|_*^c$ is an absolute value whenever $0<c\le1$ and $|x|_*$ is an absolute value over some integral domain $D$. The given statement, however, is simply a question about a real inequality. How do you prove this one?

The case $x=0$ or $y=0$ is trivial, so we can assume that $x,y>0$. Then expanding the definitions we have $e^{c\log(x+y)}\le e^{c\log x}+e^{c\log y}$, but the addition doesn't play well with the exponentials and logs, so I don't see what else can be done. This feels like a convexity result, but I'm not seeing exactly how to make the connection (not to mention that I still haven't proved that $\exp$ is convex and $\log$ is concave, so if I can avoid that I'd prefer to).

Mario Carneiro
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3 Answers3

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Here is another approach: Since the inequality is homogeneous, we can assume without loss of generality that $x+y=1$. What we then need to prove is $1\le x^c+y^c$. Since $x$ and $y$ are nonnegative, it follows then that $x\le1$ and $y\le1$. And because $0\le c\le1$ it follows that $0\le 1-c\le1$: $$x\le1 \Rightarrow x^{1-c}\le1 \Rightarrow x\le x^c$$ And the same for $y$, namely $y\le y^c$. Adding those two inequalities we get $x+y\le x^c+y^c$ and since $x+y=1$ we have: $$1\le x^c+y^c$$ If you are not familiar with homogeneity: The inequality is equivalent to: $$1\le \left(\frac{x}{x+y}\right)^c+\left(\frac{y}{x+y}\right)^c$$ If we then make the change of variables $a=\frac{x}{x+y}$ and $b=\frac{y}{x+y}$ we need to prove $1\le a^c+b^c$ with $a+b=1$. Changing back the variables as $a=x$ and $b=y$ leeds to the same conditions as with the homogeneity.

Redundant Aunt
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Consider $f(x) = (x+y)^c - x^c - y^c$ on $[0,\infty)$, we have:

$f'(x) = c(x+y)^{c-1} - cx^{c-1} = c\left((x+y)^{c-1} - x^{c-1}\right) < 0$ since $x + y > x$, and $0 \leq c < 1$. Thus $f(x) \leq f(0) = 0$. So: $(x+y)^c - x^c - y^c \leq 0$, and this means: $(x+y)^c \leq x^c + y^c$.

DeepSea
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For $q>1$, it is true that $(a^q+b^q)^{1/q}\leq a+b$ for $a,b\geq 0$ (this is the usual triangular inequality). Letting $p=1/q$, $x=a^q$ and $b=y^q$ (that is, $a=x^p$, $b=y^p$), we obtain $(x+y)^p\leq x^p+y^p$.

Luiz Cordeiro
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  • Do you know a slick way to prove the "usual triangular inequality"? Really it's just trading this problem for $(x+y)^c\ge x^c+x^c$ when $c\ge1$, which doesn't look any easier. – Mario Carneiro Sep 08 '14 at 21:49
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    Oh, right. I was assuming you already knew this. You can consider the $\ell^q$-norm, $\Vert\cdot\Vert_q$ in $\mathbb{R}^2$. Then $(a^q+b^q)^{1/q}=\Vert (a,b)\Vert=\Vert (a,0)+(0,b)\Vert_q\leq\Vert (a,0)\Vert_q+\Vert(0,b)\Vert_q=a+b$. The problem is that to show $\Vert\cdot\Vert_q$ is a norm, you have to either follow the proof of Hölder's Inequality or make an argument similar to the one Dark-Chocolate used. – Luiz Cordeiro Sep 08 '14 at 22:05
  • How would you prove this using Hölder's Inequality? – Mario Carneiro Sep 09 '14 at 07:09