Prove that If $a,b,c \in R^+$ Then
$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}} \gt 2$$
My try:
I assumed:
$$x=\sqrt{\frac{a}{b+c}}$$
$$y=\sqrt{\frac{b}{a+c}}$$
$$x=\sqrt{\frac{c}{b+a}}$$
Then we have:
$$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)$$
Now by AM GM Inequality we have:
$$\left(\frac{a}{b}+\frac{b}{a}\right) \ge 2$$
So we get:
$$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \ge 6 \tag{1}$$
Using $(1)$ we need to prove:
$$x+y+z \gt 2$$
Any idea here?
Writing cyclic sums for short, $$\sum_{cyc}\sqrt{\frac{a}{b+c}}=\sum_{cyc}\frac{2a}{2\sqrt{a(b+c)}}\\ \geq\sum_{cyc}\frac{2a}{a+b+c}=2 $$ where the AM-GM inequality was used for each denominator, i.e. $2\sqrt{a(b+c)} \le a + (b+c)$
As Martin R. pointed out, math.stackexchange.com/a/2237066/42969 has the same reasoning, so credits belong to Michael Rozenberg. Here, some extra explanation is given.