<s>[INST] Let $a,b,c$ be the lenghts of the sides of a triangle. Suppose that $ab+bc+ca=1$. Show that $(a+1)(b+1)(c+1)<4$ Let $a,b,c$ be the lenghts of the sides of a triangle. Suppose that $ab+bc+ca=1$. Show that $(a+1)(b+1)(c+1)<4$.  
My attempt:  
I tried multiplying the whole thing but that didn't help at all. So, I tried to manipulate the triangle inequality and bring out the given form but that didn't help too. I am out of ideas now. Please help. Thank you.
 [/INST] Note 
$$(a+1)(b+1)(c+1)=1+abc+a+b+c+ab+bc+ca.$$ So it suffices to show $$abc+a+b+c<2,$$
or $$abc+a+b+c-2=abc+a+b+c-1-ab-bc-ca<0.$$
This is the same as 
$$(a-1)(b-1)(c-1)<0.$$
The only problematic possibility is $a>1>b>c.$ In this case,
$$1=ab+bc+ca>b+bc+c>b+c>a,$$
a contradiction.
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