Can Minkowski's Inequality Prove Summation Inequality for Positive Numbers?

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Minkowski's Inequality can be used to demonstrate that the summation inequality (\sum x_i )^a ≤ \sum x_i^a holds for non-negative numbers x_i when 0 < a < 1. To prove this, one must establish that (x+y)^\alpha ≤ x^\alpha + y^\alpha for non-negative x and y, with 0 < α < 1. This involves analyzing the function f(x) = 1 + x^\alpha - (1 + x)^\alpha, showing that it is increasing and has a minimum at 0. The conclusion is that f(x) ≥ 0, which supports the original inequality. The discussion highlights the challenge of proving this inequality without Minkowski's framework.
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I don't understand how it is possible to show using the Minkowski's Inequality that
(\sum x_i )^a \leq \sum x_i^a where x_i \geq 0 \forall i and 0&lt;a&lt;1.

I also tried to prove this without using Minkowski, but to no avail.

This is driving me crazy although it seems to be trivial in the literature.
I will appreciate any help
 
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Hi St41n! :smile:

It seems that you must prove that (x+y)^\alpha\leq x^\alpha+y^\alpha for x,y\geq 0 and 0&lt;\alpha&lt;1.

For that, you must look at the function

f:\mathbb{R}^+\rightarrow \mathbb{R}:x\rightarrow 1+x^\alpha-(1+x)^\alpha

Try to show that f is increasing and has its minimum in 0. It follows that f(x)\geq 0.
 
Thank you very much for the quick reply!
 
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