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 loss of generality, $a \leqslant b \leqslant c = 1$. Thus it remains to see that
$$1 \leqslant \frac{1+a^2+b^2}{a + b + ab} \leqslant 2.$$
For the left inequality, we rewrite
$$a + b + ab \leqslant 1 + a^2 + b^2 \iff 0 \leqslant 1 - a - b + ab + a^2 -2ab + b^2 = (1-a)\cdot(1-b) + (b-a)^2,$$
which is evident.
For the right inequality,
$$1+a^2+b^2 \leqslant 2(a+b) + 2ab \iff 1 + (b-a)^2 \leqslant 2(a+b),$$
which is seen to be true since by assumption - that $a \leqslant b\leqslant 1$ be the sides of a triangle - $0 \leqslant b-a \leqslant 1 \leqslant a+b$.
letting $a \to 0$ resp. $a \to 1$ (both imply $b \to 1$) shows that the bounds are sharp.
