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. 
 A: let $a=x^2,b=y^2,c=z^2$
$$\Longleftrightarrow x^2+y^2+z^2=3\Longrightarrow \dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\ge 3$$
note
$$(x^2+y^2+z^2)^2=x^4+y^4+z^4+2x^2y^2+2z^2x^2+2y^2z^2$$
use
AM-GM inequality,then we have
$$4\dfrac{x^2}{y}+2x^2y^2+x^4\ge (2\cdot 2+2+1)x^{\frac{2\cdot 2\times 2+2\cdot 2+2\cdot 2}{2\cdot 2+2+1}}=7x^{\frac{16}{7}}$$
and the same
$$4\dfrac{y^2}{z}+2y^2z^2+y^4\ge 7y^{\frac{16}{7}}$$
$$4\dfrac{z^2}{x}+2z^2x^2+z^4\ge 7z^{\frac{16}{7}}$$
so
$$4\sum\dfrac{x^2}{y}+(x^2+y^2+z^2)^2\ge 7\sum x^{\frac{16}{7}}$$
$$\Longleftrightarrow x^{\frac{16}{7}}+y^{\frac{16}{7}}+z^{\frac{16}{7}}\ge x^2+y^2+z^2$$
use AM-GM inequality 
we have
$$7x^{\frac{16}{7}}+1=x^{\frac{16}{7}}+x^{\frac{16}{7}}+\cdots+x^{\frac{16}{7}}+1\ge 8\sqrt[8]{x^{\frac{16}{7}\cdot 7}}=8x^2 $$ 
$$7\sum x^{\frac{16}{7}}+\sum x^2\ge \sum 8x^2$$
so
$$\sum x^{\frac{16}{7}}\ge \sum x^2$$
In  general,we have
$x^n+y^n+z^n=3,2p+q>2n,p,q,n\in N^{+}$,then
$$\dfrac{x^p}{y^q}+\dfrac{y^p}{z^q}+\dfrac{z^p}{x^q}\ge 3$$
A: Rewrite it as 
$$\sum_{cyc} \frac{a}{\sqrt{b}} + (\sqrt{b} - 2\sqrt{a}) = \sum_{cyc} \frac{a}{\sqrt{b}} -\sum_{cyc} \sqrt{a}  \ge 3 - \sum_{cyc} \sqrt{a} = \sqrt{3(a+b+c)} - \sum_{cyc} \sqrt{a}$$
LHS becomes
$$\sum_{cyc} \frac{(\sqrt{a} - \sqrt{b})^2}{\sqrt{b}}$$
RHS becomes
$$\frac{3(a+b+c) - (\sum_{cyc} \sqrt{a})^2}{\sqrt{3(a+b+c)} + \sum_{cyc} \sqrt{a}} = \frac{\sum_{cyc} (\sqrt{a} - \sqrt{b})^2}{\sqrt{3(a+b+c)} + \sum_{cyc} \sqrt{a}}$$
The inequality then follows since
$\sqrt{3(a+b+c)} + \sum_{cyc} \sqrt{a} \ge \sqrt{a}, \sqrt{b}$ and $\sqrt{c}$.
A: 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 $2\times \{a^2+b^2+c^2-(ab+bc+ca)\} = (a-b)^2 +(b-c)^2 + (c-a)^2 \geq 0$
A: By Holder’s inequality (with $\sum$ denoting cyclic sums):
$$\left(\sum \frac{a}{\sqrt b}\right)^2\left(\sum ab \right)\geqslant (a+b+c)^3=27$$
So it is enough to show $\sum ab \leqslant 3=\frac13(a+b+c)^2$, which is well known.
A: 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.
A: 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!
A: 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.
