<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] Since $\left(1/\sqrt{x}\right)^{\prime\prime} = \frac{3}{4 x^{5/2}} > 0$ for $x > 0$, we have that $f(x)=\frac{1}{\sqrt{x}}$ is convex.
Then, by Jensen's inequality we have:
$$ \tfrac{1}{a+b+c}\times \left( \frac{a}{\sqrt{b}} + \frac{b}{\sqrt{c}} + \frac{c}{\sqrt{a}} \right) \geq \left( \frac{ab+bc+ca}{a+b+c} \right)^{-1/2} $$
Here $9=(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) \geq 3\times (ab+bc+ca)$  $(*)$, so we actually get $3\geq ab+bc+ca$ and therefore:
$$ \tfrac{1}{a+b+c}\times \left( \frac{a}{\sqrt{b}} + \frac{b}{\sqrt{c}} + \frac{c}{\sqrt{a}} \right) \geq \left( \frac{ab+bc+ca}{a+b+c} \right)^{-1/2} \geq \left( \frac{3}{a+b+c} \right)^{-1/2} = 1 $$
Since we have $a+b+c=3$, we are done.
$(*)$ This follows from $2\times \{a^2+b^2+c^2-(ab+bc+ca)\} = (a-b)^2 +(b-c)^2 + (c-a)^2 \geq 0$
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