If $a,b,c$ are positive reals and $abc=1$, Prove that $$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \geq \sqrt{a}+\sqrt{b}+\sqrt{c}$$
My try: $$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}=\frac{b+c+a-a}{\sqrt{a}}+\frac{c+a+b-b}{\sqrt{b}}+\frac{a+b+c-c}{\sqrt{c}}$$ $\implies$ $$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}=(a+b+c)(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})-(\sqrt{a}+\sqrt{b}+\sqrt{c})$$ Is there a way from here?