Let $a,b,c$ be the lenghts of the sides of a triangle. Suppose that $ab+bc+ca=1$. Show that $(a+1)(b+1)(c+1)<4$

Question

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.

Answer 1

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.

Answer 2

I could just show that it is less than 5 since I cannot comment I am just posting my partial answer.

$(a+1)(b+1)(c+1)=1+abc+a+b+c+ab+bc+ca$. So we have to just show that $abc+a+b+c<2$.

Now $a+b+c+abc \leq( a+b)(2+ab)$. Also not that $ab,bc,ca <1$. So atleast two sides must be less than $1$. wlog assume that they are $a$ and $b$. This implies $ab<a+b$. So $( a+b)(2+ab) < (a+b+1)^2-1 <3$