<s>[INST] Prove $a+b+c \geq ab+bc+ca$, given an additional constraint 
If $a,b,c$ are positive real numbers satisfying
  $$\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\geq 1$$
  then I'd like to prove that
  $$a+b+c \geq ab+bc+ca\,.$$
  Additional info:
  We should only use Cauchy-Schwarz (preferred to be used at least once) and AM-GM. We are not allowed to use induction.

Things I have tried so far: Using the Cauchy-Schwarz on the given constraint inequality, I can show
$$\left(\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\right)\big((a+b+1)+(b+c+1)+(c+a+1)\big)\geq (1+1+1)^2$$
$$\left(\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\right)(2(a+b+c)+3)\geq9$$
$$2(a+b+c)+3\geq 9$$
$$a+b+c\geq 3$$
So my idea is right now to show $3\geq ab+bc+ca$.
And trying Cauchy-Schwarz on problem statment:$$(a+b+c)(a^2b+b^2c+c^2a)\geq(ab+bc+ca)^2$$
 [/INST] Hint:Use Cauchy-Schwarz inequality
$$(a+b+c^2)(a+b+1)\ge (a+b+c)^2$$
so
$$\dfrac{1}{a+b+1}\le\dfrac{a+b+c^2}{(a+b+c)^2}$$
so
$$1\le \sum_{cyc}\dfrac{1}{a+b+1}\le\sum_{cyc}\dfrac{a+b+c^2}{(a+b+c)^2}=\dfrac{2(a+b+c)+a^2+b^2+c^2}{(a+b+c)^2}$$
so
$$a+b+c\ge ab+bc+ac$$
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