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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.

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    Maybe this helps a little http://en.wikipedia.org/wiki/Shapiro_inequality2012-03-08

2 Answers 2

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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.

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    @vikiiii: You gave a correct proof that $3$ is **an** upper bound. You did not prove that there is no cheaper upper bound, and in fact there is, namely $2$. Roughly speaking, we cannot have simultaneously $p+q\approx r$, $q+r\approx p$, and $r+q\approx p$ (all in the sense of ratio). For then $2p+2q+2r\approx p+q+r$, which is impossible. More informally, suppose, as we can, that $p+q+r=1$. If $p+q\approx r$ and $q+r\approx p$, then $p+2q+r\approx p+r$, so $q\approx 0$ and $p\approx r\approx 1/2$.2012-03-08
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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,

$\frac{3}{2}<\frac{p}{q+r}+\frac{r}{q+p}+\frac{q}{p+r}<2$