Minimum of $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}$

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.

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.

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!