<s>[INST] Inequality exercise (olympiad) For positive $a$, $b$, $c$ such that $abc=1$. Show that $$(ab+bc+ca)(a+b+c)+6\geq 5(a+b+c).$$
From the LHS, using AM-GM, we see that $(ab+bc+ca)(a+b+c)+6\geq 3(abc)^{2/3}3(abc)^{1/3}+6=15$. But clearly, 15 can be less than $5(a+b+c)$. So this technique won't work. I fail to see some other inequalities to prove this. Any other ideas/hints? Thanks
 [/INST] Lemma:  for any postive $x,y,z>0$,and $xyz=1$ we have
$$(xy+yz+xz)^2+3\ge 4xyz(x+y+z)$$
proof:since Use Schur inequality we have $$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ac(a+c)\ge 2((ab)^{3/2}+(bc)^{3/2}+(ac)^{3/2})$$
take $a'^2=a^3,b'^2=b^3,c'^2=c^3$ and Assmue that $a'b'c'=1$so we have
$$a'^2+b'^2+c'^2+3\ge 2(a'b'+b'c'+c'a'),$$
so
$$(a'+b'+c')^2+3\ge 4(a'b'+b'c'+c'a')$$
Let $a'=xy,b'=yz,c'=zx$ by done.
let $a+b+c=p,ab+bc+ac=q,abc=r=1$,your inequality equal to
$$pq+6\ge 5p$$
Use Lemma we have
$$q^2+3\ge 4p$$
$$\Longleftrightarrow p\sqrt{4p-3}+6\ge 5p,p\ge 3$$
it is clear .Because 
$$p^2(4p-3)-(5p-6)^2=4(p-3)^2(p-1)\ge 0$$
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