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 $2$.

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!
