<s>[INST] Prove that: $(a+b)(b+c)(c+a)\ge(a+1)(b+1)(c+1)$ Let $a,b,c>0$ satisfy $abc=1 $
Prove that: $$(a+b)(b+c)(c+a)\ge(a+1)(b+1)(c+1)$$
I have tried:
Let $p=a+b+c;q=ab+bc+ca;r=abc$, the problem is: $$pq-r\ge r+p+q+1 \;\;\text{or}\;\; pq-p-q\ge3 $$
I have $pq\ge 9r$, so we need to prove $$9-p-q\ge3$$ or $$p+q\le 6$$ I have $p\le \dfrac{q^2}{3r}$ so the problem is $$\dfrac{q^2}{3}+q\le6$$ or $$(q-3)(q+6)\le 0$$ but I can't prove it and maybe it isn't true, pls help me
 [/INST] Alternative solution:
Since we are given that $abc=1$, we can assume that $a=\frac x y, b=\frac y z, c=\frac z x$.
We have to prove that $$({x^2} + yz)({y^2} + zx)({z^2} + xy) \ge xyz(x+ y)(y + z)(z+ x)$$ Notice that we have
$ ({x^2} + yz)({y^2} + zx) - xy(z + x)(z + y) = z(x + y){(x - y)^2} \ge 0 $
$ \Rightarrow ({x^2} + yz)({y^2} + zx) \ge xy(z + x)(y+z) $
(Use this to complete the solution.)
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