Inequality. $\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3$ Let $a,b,c \in (0, \infty)$, with $a+b+c=3$. How can I prove that: 
$$\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3 ?$$. 
I try to use Cauchy-Schwarz rewriting the inequality like : 
$$\sum_{cyc}\frac{a\sqrt{b}}{b} \geq \frac{(\sum_{cyc}{\sqrt[4]{a^2b}})^2}{a+b+c}$$  but I don't obtain anything. 

With homogenizazion: We need to prove for all $a,b,c>0$: $$\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq\sqrt{3}\sqrt{a+b+c}.$$ By squaring this is equivalent to $$\frac{\sum_{cyc} a^3c+2\sum_{cyc} a^2 b^\frac32 c^\frac12}{abc}\geq \frac{3\sum_{cyc} a^2bc}{abc}$$ which follows immediately from AM-GM, for example for the first sum, use the cyclical versions of $$2ab^3+4a^3c+bc^3\geq 7a^2bc$$ by AM-GM.
