An inequality for sides of a triangle

Question

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?

Answer 1

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 $$(1-a)(1-b)>0\\ \implies ab>a+b-1\\ =\left\vert a+b-1\right\vert\\ \implies a^2b^2>(a+b-1)^2\\ \iff 1-a^2b^2<2(a+b)-(a+b)^2\\ \iff a+b+(1+ab)\frac{1-ab}{a+b}<2\\ \iff a+b+c +abc<2\\ \implies (1+a)(1+b)(1+c)<4.$$

Answer 2

Solving $ab+bc+ca=1$ for $c$ gives $$ c=\frac{1-ab}{a+b}\tag{1} $$ The triangle inequality says that for non-degenerate triangles $$ |a-b|\lt c\lt(a+b)\tag{2} $$ Multiply $(2)$ by $a+b$ to get $$ |a^2-b^2|\lt1-ab\lt(a+b)^2\tag{3} $$ By $(3)$, we have $(a+b)^2-1+ab\gt0$; therefore, $$ \begin{align} (a+b+1)(a+b+ab-1) &=\left[(a+b)^2-1+ab\right]+(a+b)ab\\ &\gt(a+b)ab\\[6pt] &\gt0\tag{4} \end{align} $$ Furthermore, $\color{#C00000}{(3)}$ implies $$ a\ge1\implies1-b^2\le\color{#C00000}{a^2-b^2\lt1-ab}\implies b\gt a\tag{5a} $$ Similarly, $$ b\ge1\implies1-a^2\le\color{#C00000}{b^2-a^2\lt1-ab}\implies a\gt b\tag{5b} $$ Inequalities $(5)$ imply that if either $a\ge1$ or $b\ge1$, then both $a\gt1$ and $b\gt1$. Consequently, we have both $a\gt b$ and $b\gt a$. Therefore, we must have $$ a\lt1\qquad\text{and}\qquad b\lt1\tag{6} $$ Using $(1)$, $(4)$, and $(6)$, we have $$ \begin{align} abc+a+b+c-2 &=(1+ab)\frac{1-ab}{a+b}+(a+b)-2\\ &=\frac{1-a^2b^2+(a+b)^2-2(a+b)}{a+b}\\ &=\frac{(a+b-1)^2-a^2b^2}{a+b}\\ &=\frac{(a+b+ab-1)(a+b-ab-1)}{a+b}\\ &=-\frac{(a+b+ab-1)(a-1)(b-1)}{a+b}\\ &\lt0\tag{7} \end{align} $$ Therefore, since $ab+bc+ca+1=2$, $(7)$ says $$ \begin{align} (a+1)(b+1)(c+1) &=(abc+a+b+c)+(ab+bc+ca+1)\\[4pt] &\lt2+2\\[4pt] &=4\tag{8} \end{align} $$

Answer 3

$(1+a)(1+b)(1+c) = 2 + a+b+c + abc < 2 + 2(a+b) + ab(a+b) = 2 + (a+b)(2 + ab)$ (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