Find the range of the given expression? If p , q , r denotes the side of the triangle ,then the below expression will always lies between?
$$\left(\frac{p}{q+r}\right) + \left(\frac{r}{q+p}\right) + \left(\frac{q}{p+r}\right)  $$
I tried to solve it by using the property that sum of length of 2 sides of triangle is always greater than 3rd side.
So 


p+q > r 
q+r > p
r+q >p


So the expression should be less than 3.
But the answer in the book is : Range : 1.5 to 2.
Thanks in advance.
 A: There are two inequalities to prove. One is quite easy, the other less so.
Let $s$ be the half-perimeter.  By the Triangle Inequality, all of $q+r$, $p+r$, and $p+q$ are $> s$.
It follows that 
$$\frac{p}{q+r}+\frac{q}{p+r}+\frac{r}{p+q}<\frac{p}{s}+\frac{q}{s}+\frac{r}{s}=\frac{2s}{s}=2.$$
The other inequality is contest material. It can be derived from various other popular contest inequalities, such as the Rearrangement Inequality, or the Chebyshev Inequality, even AM/GM, even less. This inequality holds for all positive $p$, $q$, and $r$.  We give a fairly minimal but tricky proof.
Our sum can be written as
$$\frac{2p+2q+2r}{2q+2r}+\frac{2p+2q+2r}{2r+2p}+\frac{2p+2q+2r}{2p+2q}-3.$$
This is
$$\frac{1}{2}\left(1+\frac{p+q}{q+r}+\frac{r+p}{q+r}+1+\frac{q+r}{r+p} +\frac{p+q}{r+p}+1+\frac{r+p}{p+q}+\frac{q+r}{p+q}\right)-3.$$
Group terms as shown below:
$$\frac{1}{2}\left(3+\left(\frac{p+q}{q+r}+\frac{q+r}{p+q}\right) + \left(\frac{q+r}{r+p}+\frac{r+p}{q+r}\right)+\left(\frac{r+p}{p+q}+\frac{p+q}{r+p}\right)\right)-3.$$
Recall that if $t$ is positive, then $t+1/t\ge 2$. This can be proved in various ways, such as noting that $(\sqrt{t}-1/\sqrt{t})^2 \ge 0$.  Each of our three groups above is of the shape $t+1/t$.
We conclude that our original expression is $\ge (1/2)(3+6)-3$, which is $3/2$.
Note that the upper bound of $2$ cannot be achieved with a genuine triangle, though we can get arbitrarily close. The lower bound is achieved in the equilateral case.
A: The expression can be simplified in another way as shown,
$$\frac{p}{q+r}+\frac{r}{q+p}+\frac{q}{p+r}$$
$$(\frac{p}{q+r}+1)+(\frac{r}{q+p}+1)+(\frac{q}{p+r}+1) - 3$$
$$(p+q+r)({\frac{1}{q+r}+\frac{1}{q+p}+\frac{1}{p+r}})-3$$
$$\frac{1}{2}(\frac{q+r}{1}+\frac{q+p}{1}+\frac{p+r}{1})({\frac{1}{q+r}+\frac{1}{q+p}+\frac{1}{p+r}})-3$$
Applying Arithmetic mean greater than Harmonic mean inequality we get 
$$(\frac{q+r}{1}+\frac{q+p}{1}+\frac{p+r}{1})({\frac{1}{q+r}+\frac{1}{q+p}+\frac{1}{p+r}})>9$$
Hence,
$$\frac{p}{q+r}+\frac{r}{q+p}+\frac{q}{p+r}>\frac{9}{2}-3$$
$$\frac{p}{q+r}+\frac{r}{q+p}+\frac{q}{p+r}>\frac{3}{2}$$
For lower bound as Andre points out in the previous answer we consider the case of an equilateral triangle,to get,
$$\frac{3}{2}<\frac{p}{q+r}+\frac{r}{q+p}+\frac{q}{p+r}<2$$
