<s>[INST] 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?
 [/INST] WOLOG assume that $a\geq b\geq c>0$. The constraints the problem imposes are $c=\frac{1-ab}{a+b}$, $b+c>a$. Equivalently, we have $\frac{1-ab}{a+b}+b > a\geq b$ and $b\geq \frac{1-ab}{a+b}$. These inequalities yield $$0\leq a^2-b^2<1-ab<1\quad (1)$$ and $$ab\geq \frac{1-b^2}{2}.\quad(2)$$ 
Note that $(1)$ implies $0<b\leq a<1$. Now, suppose that $a+b\leq 1$. Then from $(2)$ we have $2b\geq 2b(a+b)\geq 1+b^2$, which is not possible since $b<1$. Thus, we must have $$a+b>1.\quad (3)$$ Invoking $(3)$ we have 
$$(1-a)(1-b)>0\\
\implies ab>a+b-1\\
=\left\vert a+b-1\right\vert\\
\implies a^2b^2>(a+b-1)^2\\
\iff 1-a^2b^2<2(a+b)-(a+b)^2\\
\iff a+b+(1+ab)\frac{1-ab}{a+b}<2\\
\iff a+b+c +abc<2\\
\implies (1+a)(1+b)(1+c)<4.$$
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