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

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The forum discussion centers on proving the summation inequality for positive numbers using Minkowski's Inequality, specifically demonstrating that (\sum x_i )^a \leq \sum x_i^a for x_i ≥ 0 and 0 < a < 1. A key approach involves proving that (x+y)^\alpha ≤ x^\alpha + y^\alpha for x, y ≥ 0 and 0 < α < 1. The function f(x) = 1 + x^\alpha - (1 + x)^\alpha is introduced, with the goal of showing that f is increasing and has its minimum at 0, leading to the conclusion that f(x) ≥ 0.

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St41n
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I don't understand how it is possible to show using the Minkowski's Inequality that
[itex](\sum x_i )^a \leq \sum x_i^a[/itex] where [itex]x_i \geq 0 \forall i[/itex] and [itex]0<a<1[/itex].

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 [itex](x+y)^\alpha\leq x^\alpha+y^\alpha[/itex] for [itex]x,y\geq 0[/itex] and [itex]0<\alpha<1[/itex].

For that, you must look at the function

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

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

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