An inequality for sides of a triangle Let $ a, b, c $ be sides of a triangle and $ ab+bc+ca=1 $. Show
$$(a+1)(b+1)(c+1)<4 $$
I tried Ravi substitution and got a close bound, but don't know how to make it all the way to $4 $.
I am looking for a non-calculus solution (no Lagrange multipliers).
Do you know how to do it?

$(1+a)(1+b)(1+c) = 2 + a+b+c + abc < 2 + 2(a+b) + ab(a+b) = 2 + (a+b)(2 + ab)$ (because $c < a+b)$
$ab + bc + ac = ab + (a+b)c < ab + (a+b)^2$ for the same reason, that is $ab > 1 - (a+b)^2$, and $2 + ab > 3 - (a+b)^2$
Combining, we have $(1+a)(1+b)(1+c) < 2 + 3(a+b) - (a+b)^3$
The latter achieves the maximum 4 at $a+b = 1$ (we are not interested in negative $a+b$), QED
