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?
 A: Hint: The issue is that the condition "sides of a triangle," though it poses an easily stated set of constraints ("Triangle Inequality") is not easy to work with algebraically.
There is, however, a standard trick. The positive numbers $a$, $b$, and $c$ are sides of a triangle if and only if there exist positive numbers $p$, $q$, and $r$ such that $a=q+r$, $b=p+r$, $c=p+q$. 
(The numbers $p,q,r$ have a simple geometric interpretation. Draw the incircle of the triangle. This meets the sides of the triangle at $3$ points, which divide each side of the triangle into $2$ segments. These segments are equal in pairs, and $p$, $q$, and $r$ are the three lengths of these segments.)
If we substitute for $a$, $b$, and $c$ in our expression, we get fairly quickly the desired bounds.
A: 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.
A: (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.
A: Let $x=p-a$, $y=p-b$, and $z=p-c$, where $p=(a+b+c)/2$. The main expression is equal to $$F:=\frac{4(x+y+z)^2}{x^2+y^2+z^2+3(xy+yz+zx)}=\frac{4\sum x^2+8\sum yz}{\sum x^2 +3\sum yz}\\=4-\frac{4}{\sum x^2/\sum yz +3}.$$ From Cauchy-Schwarz we know that $\sum x^2 \geq \sum yz$. Thus, $F\geq 3$. Furthermore, it is obvious that $F<4$. The lower bound is attainable at $x=y=z$ (i.e., $a=b=c$) while you can only tend to the upper bound by tending two of $x$,$y$, or $z$ two zero while keeping the third fixed.
