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?

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.
