An inequality on sides of triangle 
Let $a,b,c$ denote the sides of a triangle. Then prove that the quantity $a\over b+c$$+$$b\over c+a$$+$$c\over a+b$ lies between $3\over 2$ and $2$.

My attempt : 
Let $P=a+b+c$
With a little manipulation it becomes 
$-3 + {P\over P-a} + {P\over P-b} + {P\over P-c} $
Applying $AM>HM$ for the three latter terms , I got the lower limit , but I am not able to find the upper limit.
Could someone please help with the upper limit ?
Thanks.

$$2>F=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}~~~(1)$$
$$F+3=(a+b+c)(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b})~~~~(2)$$
Using AM_HM:
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge \frac{9}{x+y+z}~~~~(3)$$
Using (3) in (2), we prove that
$$F+3 \ge \frac{9(a+b+c)}{2(a+b+c)} \implies F \ge 9/2-3= \frac{3}{2}.$$
The equality holds when the triangle is equilateral.
For LHS, in a triangle $$\frac{a}{b+c}<1 \implies \frac{a}{b+c} <\frac{2a}{a+b+c}$$
When we add the same positive number up and down in a proper fraction we get a bigger
fraction:
If $$\frac{P}{Q}<1 \implies \frac{P}{Q}< \frac{P+R}{Q+R},~~ P,Q,R>0$$
 Similarly we get two more results:
$$\frac{b}{c+a} <\frac{2b}{a+b+c},~~~\frac{c}{a+b} < \frac{2c}{a+b+c}$$
Adding last three results we get
$$F=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} <2$$
