<s>[INST] To prove $a+b+c \ge ab+bc+ca$ when $abc=1$ The actual question is not in the heading, it's this-

Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that
$$ {a \over \sqrt{7+b+c}} + {b \over \sqrt{7+c+a}} + {c \over \sqrt{7+a+b}} \geq 1$$

To complete a proof of this inequality after a deft application of Hölder's Inequality, it's enough to prove that
$$\tag{1} (a+b+c)^3\geq7(a+b+c)+2(ab+bc+ca)$$
and it turns to
$$ \tag{2} (a+b+c)^3\geq7(a+b+c)+{2\over 3}(a+b+c)^2 $$ as $ a+b+c\geq3 $.
According to my book, this solution is provided.
But I am not able to understand how ($1$) $\Rightarrow$ ($2$). Can someone explain me?
For this to happen, $a+b+c\geq ab+bc+ca$.
Thanks!
 [/INST] We need to prove
$$\tag{1} (a+b+c)^3\geq7(a+b+c)+2(ab+bc+ca).$$
But from know inequality $3(ab+bc+ca) \leqslant (a+b+c)^2.$ Therefore the proof is completed if
$$\tag{2} (a+b+c)^3\geq7(a+b+c)+\frac23(a+b+c)^2.$$
Let $x = a+b+c \geqslant 3\sqrt[3]{abc} = 3,$ ienquality $(2)$ become
$$x^3 \geqslant 7x+\frac{2x^2}{3},$$
or
$$\frac{x(3x+7)(x-3)}{3} \geqslant 0.$$
Which is true.
Note. Must be $(2) \Rightarrow (1) $
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