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I want to prove this, where $x_1,...,x_n$ are positive real numbers: $$(x_1^2+...+x_n^2)^2 \leq (x_1+...+x_n)(x_1^3+...+x_n^3)$$

I have written a proof but I am not very happy with it, using the Cauchy-Schwarz inequality ($|<x,y>| \leq \|x\|\|y\|$):

$$ (x_1^2+...+x_n^2)^2 = \langle (x_1^{\frac{1}{2}} ,..., x_n^{\frac{1}{2}}),(x_1^{\frac{3}{2}} ,..., x_n^{\frac{3}{2}}) \rangle^2 \leq \|(x_1^{\frac{1}{2}} ,..., x_n^{\frac{1}{2}})\|^2 \|(x_1^{\frac{3}{2}} ,..., x_n^{\frac{3}{2}})\|^2 = (x_1+...+x_n)(x_1^3+...+x_n^3)$$

Is there a better proof?

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3  
Why aren't you happy with your proof involving CS? In any proof, at some stage you need to use the fact that a square is non-negative, which is what CS precisely uses as well. –  user17762 Feb 2 '13 at 22:38
    
Your proof is very compact, effectively the inequality is a special form of CS. I doubt there is a shorter proof. –  sdcvvc Feb 2 '13 at 22:39
    
Your proof is great. –  1015 Feb 2 '13 at 22:39
    
Okay, thanks! I found it somewhat artificial, but I will stick with it. –  Alderz Feb 2 '13 at 22:44

1 Answer 1

up vote 2 down vote accepted

$$\sum_i x_i\sum_j x_j^3 - \sum_i x_i^2\sum_j x_j^2 = \sum_{i< j} (x_i x_j^3 + x_i^3 x_j - 2 x_i^2 x_j^2) = \sum_{i< j} x_i x_j (x_i-x_j)^2 \ge 0$$

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Very nice, thanks! –  Alderz Feb 2 '13 at 23:53
3  
@Alderz Just in case you haven't seen it before, this is a particular case of an identity of Lagrange, who used it to prove the Cauchy-Schwarz inequality. –  Andres Caicedo Feb 3 '13 at 0:30

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