Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there any bound for an expression like:

$$\left( a_1 + a_2 + \cdots + a_n\right)^{1/2} \leq ?$$

I need it for $n=3$. I know Hardy's inequality but it is for exponent greater than 1. Is there anything for the square root?

Thank you all!

share|improve this question
2  
there are many bounds, for example $$\max \{3, 3\cdot \max{a_n}\}$$ is a upper bound –  Dominic Michaelis Mar 2 '13 at 13:53
    
I'm so stupid! Of course there is.. $(a_1+a_2)^{1/2} = \|(a_1^{1/2}, a_2^{1/2})\| \leq a_1^{1/2} + a_2^{1/2}$.. the norm is less than the sum of the sides.. –  Dann Mar 2 '13 at 13:54
    
Oh! Interesting! Maybe that's better.. which one is sharper? –  Dann Mar 2 '13 at 13:54
    
yours is sharper –  Dominic Michaelis Mar 2 '13 at 13:55
3  
@Dann: You are not stupid. –  dot dot Mar 2 '13 at 14:03

1 Answer 1

Elementary proof from scratch: $$(\sqrt{a_1}+\sqrt{a_2})^2 = a_1+a_2+2\sqrt{a_1a_2}\ge a_1+a_2 $$ hence $$\sqrt{a_1+a_2}\le \sqrt{a_1}+\sqrt{a_2}$$ For general $n$, by induction: $$\sqrt{(a_1+\dots+a_{n-1})+a_n}\le \sqrt{a_1+\dots+a_{n-1}}+\sqrt{a_n} \le \sqrt{a_1}+\dots+\sqrt{a_n}$$


More generally, the function $f(x)= x^p$ is subadditive for $0<p<1$, meaning $f(a+b)\le f(a)+f(b)$. A fun way to prove this is $$ f(a+b)-f(b)=\int_b^{a+b} f'(x)\,dx = \int_0^{a} f'(x+b)\,dx\le \int_0^{a} f'(x)\,dx = f(a) $$ where the inequality holds because $f'$ is decreasing.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.