<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] Assume that $x,y,z\in\mathbb{R}^+$ and $x^2+y^2+z^2=3$. We want to prove:
$$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq 3.$$
Since $f(w)=\frac{1}{w}$ is a convex function on $\mathbb{R}^+$, we have:
$$ \frac{x^2}{3}f(y)+\frac{y^2}{3}f(z)+\frac{z^2}{3}f(x)\geq f\left(\frac{x^2y+y^2 z+z^2 x}{3}\right), $$
so:
$$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq \frac{9}{x^2y+y^2 z+z^2 x},$$
and it is sufficient to prove $x^2y+y^2z+z^2x\leq 3$. In virtue of the Cauchy-Schwarz inequality,
$$x^2 y + y^2 z + z^2 y \leq \sqrt{(x^2+y^2+z^2)(x^2y^2+y^2z^2+z^2y^2)},$$
so it is sufficient to prove $x^2y^2+y^2z^2+z^2y^2\leq 3$, that is equivalent to $x^4+y^4+z^4\geq 3$, that is trivial in virtue of the Cauchy-Schwarz inequality, again.
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