<s>[INST] 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.
 [/INST] If $abc = 0,$ example $c = 0$ then
$${\sqrt{\frac{a(b+c)}{b^2+c^2}}} +{\sqrt{\frac{b(c+a)}{c^2+a^2}}} +{\sqrt{\frac{c(b+a)}{b^2+a^2}}} =\sqrt{\frac ab}+\sqrt{\frac ba} \geqslant 2.$$
Equality occur when $a=b,\,c=0.$
If $abc > 0,$ using the AM-GM inequality, we have
$${\sqrt{\frac{a(b+c)}{b^2+c^2}}} > {\sqrt{\frac{a(b+c)}{(b+c)^2}}} = \sqrt{\frac{a}{b+c}}=\frac{a}{\sqrt{a(b+c)}} > \frac{2a}{a+b+c}.$$
Therefore
$${\sqrt{\frac{a(b+c)}{b^2+c^2}}} +{\sqrt{\frac{b(c+a)}{c^2+a^2}}} +{\sqrt{\frac{c(b+a)}{b^2+a^2}}} > \frac{2(a+b+c)}{a+b+c}=2.$$
So a minimal value is $2.$
</s>