square root of a sum? Bound?

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?

Solution 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.