Minimum of $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}$ What is the minimum of $$f(a,b,c):=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}$$ where $a,b,c$ are positive real numbers?
When $a=b=c$, we have $f(a,b,c)=\dfrac{3}{\sqrt{2}}\approx 2.12$
When $a=1,b=c\rightarrow\infty$, we have $f(a,b,c)\rightarrow 2$. So the minimum is at most $2$.

Following mookid's hint, we can also avoid the use of Lagrange multiplicators. Normalize so that $a+b+c=1$, and then use the inequality $\sqrt{\dfrac{a}{1-a}}\geq 2a$. This is equivalent to $a(2a-1)^2\geq 0$.
Hence $f(a,b,c)\geq 2(a+b+c)=2$. Equality cannot hold, since $a=b=c=\dfrac{1}{2}$ doesn't satisfy $a+b+c=1$. But $f(a,b,c)$ can arbitrary get close to $2$, as the example in the original question shows.
