Limit on the expression containing sides of a triangle To find the bounds of the expression $\frac{(a+b+c)^2}{ab+bc+ca}$, when a ,b, c are the sides of the triangle.
I could disintegrate the given expression as $$\dfrac{a^2+b^2+c^2}{ab+bc+ca} + 2$$  and in case of equilateral triangle, the limit is 3.
Now how to proceed further?

(Without using the condition that $a, b, c$ are sides of a triangle) We know that 
$$(a-b)^2+(b-c)^2+(c-a)^2 \geq 0 $$
This implies that $a^2 + b^2 + c^2 \geq ab + bc + ca$. Hence $\frac{a^2+b^2+c^2} { ab+bc+ca} \geq 1$, so the minimum of the initial expression is 3.

To find the maximum, you will need to use the triangle inequality, which states that $a + b \geq c $.
It might be easier for you to try and find what the limiting case is.
