Inequality with a, b, c about finding minimal and maximal value

Question

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.

Answer 1

For $c=0$ and $a=b=1$ we obtain a value $2$.

We'll prove that it's a minimal value.

Indeed, by Holder: $$\left(\sum_{cyc}\sqrt{\frac{a(b+c)}{b^2+c^2}}\right)^2\sum_{cyc}\frac{a^2(b^2+c^2)}{b+c}\geq(a+b+c)^3.$$ Thus, it's enough to prove that: $$(a+b+c)^3\geq4\sum_{cyc}\frac{a^2(b^2+c^2)}{b+c},$$ which is obvious after full expanding: $$\sum_{sym}(a^5b-a^3b^3+5a^4bc+11a^3b^2c+5a^2b^2c^2)\geq0.$$

The maximal value does not exist

because for $b=c=1$ and $a\rightarrow+\infty$ our expression close to $+\infty$.

Answer 2

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 $2.$

Answer 3

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!