<s>[INST] Let $a,b,c>0$ and $ab+bc+ca=3$. Prove that $abc(a+b)(b+c)(c+a)\leq8$ 
Let $a,b,c>0$ and $ab+bc+ca=3$. Prove that $abc(a+b)(b+c)(c+a)\leq8$

My try: $a^3b^3c^3(a+b)(b+c)(c+a)\leq8a^2b^2c^2$ so $(a^2bc+ab^2c)(ab^2c+abc^2)(abc^2+a^2bc)\leq 8ab.bc.ca$. Then choose $ab=x,bc=y,ca=z$. Problem become $x,y,z>0$, $x+y+z=3$. Prove that $(x+y)(y+z)(z+x)\leq8xyz$. It's so wrong.
So I think maybe this inequality is wrong, I try $a=\dfrac{1}{2},b=1,c=\dfrac{5}{3}$ and it still right. Can you help me for any ideas? Thank you!
 [/INST] Your idea is briliant indeed. Put $x = ab, y = bc, z = ac$, then $x+y+z = 3$, and we prove: $(x+y)(y+z)(z+x)\le 8$. By AM-GM inequality: $(x+y)(y+z)(z+x) \le \left(\dfrac{x+y+y+z+z+x}{3}\right)^3= 2^3 = 8$. Equality occurs when $x+y=y+z=z+x \implies x = y = z = 1\implies a = b =c = 1$.
</s>