Inequality with a, b, c about finding minimal and maximal value Find the minimal and maximal value (if they exist) of ${\sqrt{\frac{a(b+c)}{b^2+c^2}}} +{\sqrt{\frac{b(a+c)}{a^2+c^2}}} +{\sqrt{\frac{c(b+a)}{b^2+a^2}}}$ if  are non-negative real numbers, such that at least two of them are positive.
My attempts for the case where all variables are positive:
I tried applying AM-GM on ${\sqrt{a(b+c)}} $ and etc. and then applying $b^2+c^2=>(b+c)^2/2$,but the inequality I received was false. I also tried rewriting $a(b+c)/(b^2+c^2)=(b/a+c/a)/((b/a)^2+(c/a)^2)$ and etc. and then letting $a/b=x$, $b/c=y$ and $c/a=z$ and etc but I was stuck from there. I also tried applying Cauchy Schwarz by squaring the whole expression.

We can prove that $2$ is a minimum value also by AM-GM:
$$\sum_{cyc}\sqrt{\frac{a(b+c)}{b^2+c^2}}=\sum_{cyc}\frac{2a(b+c)}{2\sqrt{a(b+c)\cdot(b^2+c^2)}}\geq\sum_{cyc}\frac{2a(b+c)}{a(b+c)+b^2+c^2}.$$
Id est, it's enough to prove that:
$$\sum_{cyc}\frac{a(b+c)}{a(b+c)+b^2+c^2}\geq1$$ or
$$abc\sum_{cyc}\left(a^3+2a^2b+2a^2c+\frac{1}{3}abc\right)\geq0.$$
Done!
