Prove $a+b+c \geq ab+bc+ca$, given an additional constraint

Question

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:$$(a+b+c)(a^2b+b^2c+c^2a)\geq(ab+bc+ca)^2$$

Answer 1

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 $$a+b+c\ge ab+bc+ac$$