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.

  • $\begingroup$ Lies between what? $\endgroup$ – Pedro Tamaroff Mar 8 '12 at 6:52
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    $\begingroup$ @PeterT we have to find the range. $\endgroup$ – vikiiii Mar 8 '12 at 7:11
  • $\begingroup$ Is the third term supposed to be $$\frac{q}{p+r}\;?$$ $\endgroup$ – Brian M. Scott Mar 8 '12 at 7:16
  • $\begingroup$ @BrianM.Scott ya.thanks.i have edited the question. $\endgroup$ – vikiiii Mar 8 '12 at 7:21
  • $\begingroup$ Maybe this helps a little en.wikipedia.org/wiki/Shapiro_inequality $\endgroup$ – Gerenuk Mar 8 '12 at 11:55

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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  • $\begingroup$ thanks..but according to my solution using triangle property , upper bound is coming 3. what is wrong in my solution? $\endgroup$ – vikiiii Mar 8 '12 at 17:43
  • $\begingroup$ @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 $p\approx r\approx 1/2$. $\endgroup$ – André Nicolas Mar 8 '12 at 17:55

The expression can be simplified in another way as shown, $$\frac{p}{q+r}+\frac{r}{q+p}+\frac{q}{p+r}$$

$$(\frac{p}{q+r}+1)+(\frac{r}{q+p}+1)+(\frac{q}{p+r}+1) - 3$$



Applying Arithmetic mean greater than Harmonic mean inequality we get





For lower bound as Andre points out in the previous answer we consider the case of an equilateral triangle,to get,


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