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Is there any upper bound for an expression like:

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

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?

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    $\begingroup$ there are many bounds, for example $$\max \{3, 3\cdot \max{a_n}\}$$ is a upper bound $\endgroup$ – Dominic Michaelis Mar 2 '13 at 13:53
  • $\begingroup$ 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.. $\endgroup$ – Dann Mar 2 '13 at 13:54
  • $\begingroup$ Oh! Interesting! Maybe that's better.. which one is sharper? $\endgroup$ – Dann Mar 2 '13 at 13:54
  • $\begingroup$ yours is sharper $\endgroup$ – Dominic Michaelis Mar 2 '13 at 13:55
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    $\begingroup$ @Dann: You are not stupid. $\endgroup$ – dot dot Mar 2 '13 at 14:03
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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.

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    $\begingroup$ And what if, say, $a_1 < 0$? How does this affect your elementary proof in the first section? $\endgroup$ – Confounded Jul 10 at 11:30

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