Prove: $a,b,c > 0 \Rightarrow \frac{b+c}{\sqrt{a^2 + bc}} + \frac{a+c}{\sqrt{b^2 + ac}} + \frac{a+b}{\sqrt{c^2 + ab}} \geq 4$
I was able to obtain a loose bound. My method: \begin{align} \frac{b+c}{\sqrt{a^2 + bc}} + \frac{a+c}{\sqrt{b^2 + ac}} + \frac{a+b}{\sqrt{c^2 + ab}} &\geq 3\left(\frac{(b+c)(a+c)(a+b)}{\sqrt{(a^2 + bc)(b^2 + ac)(c^2 + ab)}}\right)^{\frac{1}{3}} \\ &> 3\cdot2^{\frac{1}{3}} \sim 3.78 \end{align} The first step follows from the AM-GM inequality. The second step follows from the inequality: $4(a^2+bc)(b^2+ac)(c^2+ab) \leq [(b+c)(a+c)(b+a)]^2 $ which is an inequality which has been asked as a question previously. (I am unable to find it's link somehow). Also the second step has a '$>$' instead of a '$\geq$' because both inequalities do not hold simultaneously.
How to prove the tight bound?
[I had tried Jensen's inequality to $\frac{1}{\sqrt{x}}$ and C-S, but it didn't help.]
By Holder and Schur $$\sum_{cyc}\frac{b+c}{\sqrt{a^2+bc}}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{b+c}{\sqrt{a^2+bc}}\right)^2\sum\limits_{cyc}(b+c)(a^2+bc)}{\sum\limits_{cyc}(b+c)(a^2+bc)}}\geq$$ $$\geq\sqrt{\frac{\left(\sum\limits_{cyc}(b+c)\right)^3}{\sum\limits_{cyc}(b+c)(a^2+bc)}}=\sqrt{\frac{8(a+b+c)^3}{2\sum\limits_{cyc}(a^2b+a^2c)}}=$$ $$=2\sqrt{\frac{\sum\limits_{cyc}(a^3+3a^2b+3a^2c+2abc)}{\sum\limits_{cyc}(a^2b+a^2c)}}\geq2\sqrt{\frac{\sum\limits_{cyc}(4a^2b+4a^2c+abc)}{\sum\limits_{cyc}(a^2b+a^2c)}}\geq4.$$ I used the following Schur.
For positives $a$, $b$ and $c$ prove that $$\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0.$$ In our case $$\sum_{cyc}(a^3+3a^2b+3a^2c+2abc)=$$ $$=\sum_{cyc}(a^3-a^2b-a^2c+abc+4a^2b+4a^2c+abc)\geq$$ $$\geq\sum_{cyc}(4a^2b+4a^2c+abc).$$