I have attempted this question but I believe there is a better way of doing this. Please leave some comment on how I should improve my answer/ alternative ways of doing it. Thanks!
Question:
Let $a, b, c$ be the lengths of the sides of a triangle. Suppose that $ab + bc + ca = 1$. Show that $(a+1)(b+1)(c+1)< 4$.
My attempt:
Sides of the triangle add up to a perimeter $x : a+b+c=x$
Equivalently, $(a+1) + (b+1) + (c+1) = x + 3$
By triangle inequality, assuming $a+b>c$. Equivalently, $(a+1)+(b+1) > c+1$.
For the case $(a+1) + (b+1) + (c+1)$, since the addition of the sides of the triangle are less than the sides of the triangles multiplied, we are sure that if $(a+1)(b+1)(c+1)< 4$, with our assumption above we can safely assume that $(a+1) + (b+1) + (c+1)<4 $ :
$(a+1) + (b+1) + (c+1) = x+3< 4$ or $x+3<4$.
We now know that $x\leq 1$ (the perimeter of the sides $a,b,c$ are lesser than 1), hence we know that for all value of $a,b,c \leq 1$ the statement holds.