<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] The homogeneous form of the inequality is
$$(a+b+c)^2(ca^2+ab^2+bc^2)^2\geq 27(a^2+b^2+c^2)\,a^2 b^2 c^2 $$
that is equivalent to:
$$ (a^2+b^2+c^2+2ab+2bc+2ac)(c^2 a^4+a^2 b^4 + b^2 c^4 + 2 a^3 b^2 c+2 b^3 c^2 a+2 c^3 a^2 b) \geq 27(a^4 b^2 c^2+a^2 b^4 c^2 + a^2 b^2 c^4) $$
that can be proved by combining Schur's inequality with Muirhead's inequality.
</s>