Help in proving this inequality

Question

$a,b,c \in R^+: abc=1$

Prove that $1+\frac{3}{a+b+c}\ge\frac{6}{ab+bc+ca}$

I tried replacing the RHS using GM-HM

To P.T

$LHS\ge 2GM(ab,bc,ca)=2$

But this means that $3 \ge a+b+c$

which is clealy false by AM-GM

How to proceed I don't know.

I tried assigning $a,b,c$ as the roots of a cubic but this also didn't work

Answer 1

By AM-GM $$1+\frac{3}{a+b+c}\geq2\sqrt{\frac{3}{a+b+c}}.$$ Id est, it's enough to prove that $$2\sqrt{\frac{3}{a+b+c}}\geq\frac{6}{ab+ac+bc}$$ or $$(ab+ac+bc)^2\geq3(a+b+c)abc$$ or $$\sum_{cyc}c^2(a-b)^2\geq0.$$