For all $a,b,c\ge 0: ab+bc+ca>0$ then prove $$\color{black}{\sqrt{\frac{a}{b^2+bc+c^2}}+\sqrt{\frac{b}{c^2+ca+a^2}}+\sqrt{\frac{c}{a^2+ab+b^2}}\ge 2\sqrt{2}\sqrt{\frac{(ab+bc+ca)}{(a+b)(b+c)(c+a)}}.}$$ Equality holds at $a=b=c>0$ or $a=b>0; c=0$ and permutations.
Here is my proof.
We can use Holder inequality as $$\left(\sum_{cyc}\sqrt{\frac{a}{b^2+bc+c^2}}\right)^2\cdot \sum_{cyc}a^2(b^2+bc+c^2)\ge (a+b+c)^3$$ and we need to prove $$(a+b+c)^3(a+b)(b+c)(c+a)\ge 8(ab+bc+ca)\sum_{cyc}a^2(b^2+bc+c^2).$$ Let $a+b+c=p=1; 0\le ab+bc+ca=q\le \frac{1}{3}; abc=r$ it becomes $$q-r\ge 8q[r+2(q^2-2r)] \iff r(24q-1)\ge q(4q-1)(4q+1) \tag{1}$$ My strategy is divide it into two cases.
Case $1.$ $ 24q\ge 1.$
By using Schur $9r\ge 4q-1$, we obtain \begin{align*} r(24q-1)- q(4q-1)(4q+1)\ge \frac{4q-1}{9}(24q-1)- q(4q-1)(4q+1)=\frac{(4q-1)(1-3q)(12q-1)}{9}\ge 0, \end{align*} which is true when $\dfrac{1}{4}\le q\le \dfrac{1}{3}.$
And in subcase $\dfrac{1}{24}\le q <\dfrac{1}{4},$ $(1)$ is obviously true.
Case $2.$ $ 24q\le 1.$
I use $q\ge 9r \implies r(24q-1)\ge \frac{24q^2-q}{9}.$
Thus, we obtain $$r(24q-1)- q(4q-1)(4q+1)\ge \frac{q}{9}(24q-1)- q(4q-1)(4q+1)=\frac{8q(6q+1)(1-3q)}{9}\ge 0,$$ which is true when $0\le q\le \dfrac{1}{3}.$
Hence, we solved all the cases and the proof is complete.
I'd like to ask if there is other ideas like AM-GM or Cauchy-Schwarz.
Hope you can share it to us. Thanks for your interest.
By C-S and AM-GM we obtain: $$\sum_{cyc}\sqrt{\frac{a}{b^2+bc+c^2}}=\sqrt{\sum_{cyc}\left(\frac{a}{b^2+bc+c^2}+2\frac{\sqrt{ab(b^2+bc+c^2)(a^2+ac+c^2)}}{(b^2+bc+c^2)(a^2+ac+c^2)}\right)}=$$ $$=\sqrt{\sum_{cyc}\left(\frac{a}{b^2+bc+c^2}+\frac{2\sqrt{ab(b^2+bc+c^2)(a^2+ac+c^2)}}{(b^2+bc+c^2)(a^2+ac+c^2)}\right)}\geq$$ $$\geq\sqrt{\sum_{cyc}\left(\frac{a}{b^2+bc+c^2}+\frac{\frac{4ab}{a+b}\left(ab+\frac{c(a+b)}{2}+c^2\right)}{(b^2+bc+c^2)(a^2+ac+c^2)}\right)}$$ and it's enough to prove that:$$\sum_{cyc}\left(\frac{a}{b^2+bc+c^2}+\frac{2ab(2ab+c(a+b)+2c^2)}{(a+b)(b^2+bc+c^2)(a^2+ac+c^2)}\right)\geq\frac{8(ab+ac+bc)}{(a+b)(a+c)(b+c)},$$ which is true by $uvw$ (it's $f(w^3)\geq0,$ where $f$ is a concave function).
Can you end it now?