<s>[INST] Let a,b,c be positive real numbers numbers such that $ a^2 + b^2 + c^2 = 3 $ Let $a,b,c\in\mathbb{R^+}$ such that $  a^2 + b^2 + c^2 = 3 $. Prove that 
$$ 
(a+b+c)(a/b + b/c + c/a) \geq 9. 
$$
My Attempt 
I tried AM-GM on the symmetric expression so the $a+b+c \geq 3$, but I found $a+b+c \leq 3$.
 [/INST] Consider the following optimization problem 
$$\min_{x,y,z}\{(x+y+z)(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})\}$$
subject to 
$$x^2+y^2+z^2=3$$
All we need to show is that 
$$\min_{x,y,z}\{(x+y+z)(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})\}=9$$
under the condition 
$$x^2+y^2+z^2=3$$
Set up the Lagrangian 
$$\mathcal{L}(x,y,z,\lambda)=(x+y+z)(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})+\lambda(3-x^2-y^2-z^2)$$
where $\lambda>0$. 
First order conditions wrt $x,y,z$ yield
$$\mathcal{L}_x=0\Leftrightarrow(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})+(x+y+z)(\frac{1}{y}-\frac{z}{x^2})=2\lambda x$$
$$\mathcal{L}_y=0\Leftrightarrow(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})+(x+y+z)(\frac{1}{z}-\frac{x}{y^2})=2\lambda y$$
$$\mathcal{L}_z=0\Leftrightarrow(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})+(x+y+z)(\frac{1}{x}-\frac{y}{z^2})=2\lambda z$$
Exploiting the symmetry of the system of simultaneous equations above one obtains 
$$x^*=y^*=z^*$$ 
where $(x^*,y^*,z^*)$ is the solution to the minimization problem. Invoking here the constraint we get
$$(x^*)^2+(y^*)^2+(z^*)^2=9\Rightarrow 3(x^*)^2=9\Rightarrow x^*=1$$
($x^*=1$,since $x^*\geq0$).
So $(x^*,y^*,z^*)=(1,1,1)$ minimizes the objective function i.e.
$$\min_{x,y,z}\{(x+y+z)(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})\}=(x^*+y^*+z^*)(\frac{x^*}{y^*}+\frac{y^*}{z^*}+\frac{z^*}{x^*})=(1+1+1)(\frac{1}{1}+\frac{1}{1}+\frac{1}{1})=9$$
</s>