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 $2$.
 A: Hint: this is also 
$$
\min_{a,b,c\ge 0, a+b+c=1} \sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}
=\min_{a,b,c\ge 0, a+b+c=1} \sqrt{\dfrac{a}{1-a}}+\sqrt{\dfrac{b}{1-b}}+\sqrt{\dfrac{c}{1-c}}
$$And then you can for instance use the Lagrange multiplicators.
A: If $a=b=1$ and $c\rightarrow0^+$ we get $f(a,b,c)\rightarrow2$.
We'll prove that
$$\sum_{cyc}\sqrt{\frac{a}{b+c}}\geq2.$$
Indeed, by AM-GM
$$\sum_{cyc}\sqrt{\frac{a}{b+c}}=\sum_{cyc}\frac{2a}{2\sqrt{a(b+c)}}\geq\sum_{cyc}\frac{2a}{a+b+c}=2.$$
Done!
Also we can use Holder:
$$\left( \sum_{cyc}\sqrt{\frac{a}{b+c}}\right)^2\sum_{cyc}a^2(b+c)\geq(a+b+c)^3.$$
Thus, it remains to prove that
$$(a+b+c)^3\geq4\sum_{cyc}(a^2b+a^2c)$$ or
$$\sum_{cyc}(a^3-a^2b-a^2c+2abc)\geq0,$$
which follows from Schur.
Done!
A: 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.
