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I am looking for a proof that this inequality: $$(W_1+W_2+\cdots+W_n)^a \leq W_1^a +\cdots + W_n^a$$ is valid.

I have a power function $f(W)=W^a$ where $a$ is a real number, constant but usually $0\lt a\lt 1$. The $W_i$ are real positive numbers.

Then I want to check if $\left(\sum W_i\right)^a \leq \sum(W_i^a)$

For example for $n=2$ : $(x+y)^a \leq x^a + y^a$.


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Note, that this imply $\ell^p\subset\ell^1$ for $0<p<1$ and in fact $\ell^p\subset\ell^q$ for $0<p<q<\infty$. – AD. Nov 25 '10 at 17:54

This is equivalent to proving $(\frac{W_1}{W_1+W_2+...+W_n})^a+(\frac{W_2}{W_1+W_2+...+W_n})^a+...+(\frac{W_n}{W_1+W_2+...+W_n})^a\geq 1$. Write $a_i=\frac{W_1}{W_1+W_2+...+W_n}$. Then, $a_1+a_2+...+a_n=1$, $0< a_1,a_2,...,a_n,a < 1$. So, $a_i^a > a_i$. So, $a_1^a+...+a_n^a> a_1+a_2+...+a_n=1$. QED.

Edit: I think the inequality should be strict since $a<1$.

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I like this a lot. – Jonas Meyer Nov 25 '10 at 16:40
Nice elegant proof – user17762 Nov 25 '10 at 21:28
thanks - indeed, very elegant proof. The inequality should be strict and it is valid for a>1 not just o<a<1, then. I was wondering if i can get a generalization. I mean it works for sure for the power function. So I posted this :… – Antonis Nov 25 '10 at 23:45
@Antonis: Thanks. I don't think the inequality is valid for $a>1$. Take $W_1=W_2=0.5,a=2$. Then, LHS=$1$ and RHS=$0.5$. – Timothy Wagner Nov 25 '10 at 23:51

To show that $(w_1+w_2)^a\leq w_1^a+w_2^a$ whenever $w_1$ and $w_2$ are positive, we can divide by $w_1^a$ on both sides and rewrite $\frac{w_2}{w_1}$ as $t$ to get the equivalent statement $(1+t)^a\leq 1+ t^a$ when $t\geq 0$. Notice that the two sides agree when $t$ goes to zero, and the derivative of the right-hand side with respect to $t$ is always bigger than that of the left-hand side (because $0\lt a\lt 1$), so the inequality follows.

The case of general $n$ follows by induction, using the $n=2$ case for the inductive step.

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See my answer to Please explain inequality $|x^p-y^p|\le |x-y|^p$ - the inequality you ask for follows by induction.

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