<s>[INST] 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. 
 [/INST] We need to prove that
$$\sum_{cyc}\frac{a^2}{b}\geq3,$$ where $a$, $b$ and $c$ are positives such that $a^2+b^2+c^2=3.$
Indeed,
$$\sum_{cyc}\frac{a^2}{b}-3=\sum_{cyc}\left(\frac{a^2}{b}-2a+b\right)-\left(3-a-b-c\right)=$$
$$=\sum_{cyc}\frac{(a-b)^2}{b}-\frac{9-(a+b+c)^2}{3+a+b+c}=\sum_{cyc}\frac{(a-b)^2}{b}-\frac{3(a^2+b^2+c^2)-(a+b+c)^2}{3+a+b+c}=$$
$$\sum_{cyc}\frac{(a-b)^2}{b}-\frac{\sum\limits_{cyc}(a-b)^2}{3+a+b+c}=\sum_{cyc}\frac{(a-b)^2(3+a+c)}{b(3+a+b+c)}\geq0$$ and we are done!
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