Christyan's rational expression

Note that $uv+w=uv-u-v+1=(u-1)(v-1)$ . Thus, the desired sum is $\frac{1}{(u-1)(v-1)}+\frac{1}{(v-1)(w-1)}+\frac{1}{(w-1)(u-1)}$ $=\frac{u-1+v-1+w-1}{(u-1)(v-1)(w-1)}$ $=\frac{u+v+w-3}{uvw-uv-vw-wu+u+v+w-1}$ $=\frac{-2}{1-uv-vw-wu}.$ From the identity $(u+v+w)^2=u^2+v^2+w^2+2(uv+vw+wu),$ it follows that $uv+vw+vu=\frac{1^2-2013}{2}=-1006.$ Thus, the sum is equal to $-\frac{2}{1007}$ , so the numerator is equal to $\boxed{2}$ .

Squaring the equation $u + v + w = 1$ , we get

$u^2 + v^2 + w^2 + 2uv + 2uw + 2vw = 1.$

Then $2013 + 2(uv + uw + vw) = 1$ , so $uv + uw + vw = -1006$ .

We can then write $\begin{aligned} &\frac{1}{uv + w} + \frac{1}{wu + v} + \frac{1}{vw + u} \\ &= \frac{1}{uv + w(u + v + w)} + \frac{1}{wu + v(u + v + w)} + \frac{1}{vw + u(u + v + w)} \\ &= \frac{1}{uv + uw + vw + w^2} + \frac{1}{uv + uw + v^2 + vw} + \frac{1}{u^2 + uv + uw + vw} \\ &= \frac{1}{(u + w)(v + w)} + \frac{1}{(u + v)(v + w)} + \frac{1}{(u + v)(u + w)} \\ &= \frac{(u + v) + (u + w) + (v + w)}{(u + v)(u + w)(v + w)} \\ &= \frac{2(u + v + w)}{(u + v)(u + w)(v + w)}. \end{aligned}$

Since $u + v + w = 1$ , this is equal to $\begin{aligned} &\frac{2}{(1 - w)(1 - v)(1 - u)} \\ &= \frac{2}{1 - (u + v + w) + (uv + uw + vw) - uvw} \\ &= \frac{2}{1 - 1 - 1006 - 1} \\ &= -\frac{2}{1007}. \end{aligned}$

The final answer is 2.

(u+v+w)^2= u^2+v^2+w^2 +2(uv+vw+wu) and we get uv+vw+wu=-1006 u=1-v-w v=1-u-w w=1-u-v so, we get the given expression as 1/(u-1)(v-1) + 1/(v-1)(w-1) + 1/(w-1)(u-1) which is equal to (u-1+v-1+w-1)/(u-1)(v-1)(w-1) = (u+v+w-3)/(uvw-1-uv-vw-wu+u+w+v)= -2/1007 by substituting the values and hence a=2

Let $$S = \sum {\text{cyc}} \frac{1}{uv + w}$$ Now because $uvw = 1$ we get $$S = \sum {\text{cyc}} \frac{1}{\frac{1}{u} + u} = \sum {\text{cyc}} \frac{u}{u^2 + 1}$$ Adding the fractions we get $$S = \bigg(\sum {\text{sym}} u^2v + \sum {\text{cyc}} u^2v^2w + \sum {\text{cyc}} u\bigg)\Bigg/ \bigg(u^2v^2w^2 + \sum {\text{cyc}} u^2v^2 + \sum {\text{cyc}} u^2 + 1\bigg) = \frac{A}{B}$$ Let's compute $A$ and $B$ individually. Now $$\left(\sum {\text{cyc}} u\right)^2 = 1 = \sum {\text{cyc}} (u^2 + 2uv) = 2013 + 2\sum {\text{cyc}} uv$$ Hence $$\sum {\text{cyc}} uv = -1006$$ Now $$\sum {\text{cyc}} u^2v^2w = uvw\sum {\text{cyc}} uv = -1006$$ $$\sum {\text{sym}} u^2v = \left(\sum {\text{cyc}}u\right)\left(\sum {\text{cyc}} uv\right) - 3uvw = 1 \cdot (-1006) - 3 = -1009$$ So $$A = -1009 - 1006 + 1 = -2014$$ Also $$\left(\sum {\text{cyc}} uv\right)^2 = (-1006)^2 = \sum {\text{cyc}} u^2v^2 + 2uvw\sum {\text{cyc}}u = \sum {\text{cyc}}u^2v^2 + 2$$ Hence $$\sum {\text{cyc}} u^2v^2 = (-1006)^2 - 2 = 1012034$$ Hence $$B = 1 + 1012034 + 2013 + 1 = 1014049$$ Hence $$-\frac{a}{b} = \frac{A}{B} = \frac{2014}{1014049} = \frac{2}{1007}$$ So $$a = \boxed{2}$$

Notice that $\begin{aligned} u &=& 1-v-w \\ v &=& 1-u-w \\ w &=& 1-u-v \end{aligned}$

Subtitute these values. $\frac{1}{uv+w}+\frac{1}{wu+v}+\frac{1}{vw+u}$ $\begin{aligned} &=& \frac{1}{uv+(1-u-v)} + \frac{1}{wu+(1-u-w)} + \frac{1}{vw+(1-v-w)} \\ &=& \frac{1}{(u-1)(v-1)}+\frac{1}{(w-1)(u-1)}+\frac{1}{(v-1)(w-1)} \\ &=& \frac{(w-1)+(v-1)+(u-1)}{(u-1)(v-1)(w-1)} \\ &=& \frac{(u+w+v)-3}{uvw-uv-uw-vw+u+v+w-1} \\ &=& \frac{1-3}{uvw-(uv+uw+vw)+(u+v+w)-1} \\ &=& \frac{-2}{1-(uv+uw+vw)+1-1} \\ \end{aligned}$

How to find $uv+uw+vw$ ? Use the equations given in the problem! $u+v+w=uvw=1$ $u^2+v^2+w^2=2013$

Square $u+v+w$ , and see what would we have.

$\begin{aligned} (u+v+w)^2 &=& u^2+v^2+w^2+2uv+2uw+2vw \\ 1^2 &=& (u^2+v^2+w^2)+2(uv+uw+vw) \\ 1 &=& 2013+ 2(uv+uw+vw) \\ \end{aligned}$

We get $uv+uw+vw=\frac{1-2013}{2}=-1006$ .

Now, let's return to the equation we left. $\begin{aligned} \frac{-2}{1-(uv+uw+vw)+1-1} &=& \frac{-2}{1-(-1006)+1-1} \\ &=& \frac{-2}{1007}\\ \end{aligned}$

So, our answer is $\boxed{2}$ .

$uvw = 1$ $u + v + w = 1$

$\Rightarrow u^2 + v^2 + w^2 +2\sum uv = 1 \Rightarrow \sum uv = -1006.$

Now ,

$uv + w = uv + 1 - u - v = (u - 1)(v - 1)$ .

Similarly,

$wu + v = (w - 1)(u - 1)$ , and

$vw + u = (v - 1)(w - 1)$

Now ,

$\frac{1}{uv + w} + \frac{1}{wu + v} + \frac{1}{vw + u}$

$= \frac{1}{(u - 1)(v - 1)} + \frac{1}{(w - 1)(u - 1)} +\frac{1}{(v - 1)(w - 1)}$

$= \frac{u + v + w - 3}{(u - 1)(v - 1)(w - 1)} = \frac{ -2 }{(u - 1)(v - 1)(w - 1)}$

$(u - 1)(v - 1)(w - 1) = uvw -( u + v + w ) - \sum uv - 1$

= 2007).

Hence , the sum becomes $\frac{-2}{2007}$ and hence $a =\fbox{2}$

Note : I found a small mistake in the question , $u , v$ and $w$ can not be all reals. They are roots of equation :

$x^3 - x^2 - 1006x - 1 = 0$ , which has only one real root .

Algebra is a place where some expressions fit in miraculously better than others.

Observe that: $\frac{1}{u v + w} = \frac{1}{u v + 1 - u - v} = \frac{1}{(u-1)(v-1)}$ and the like. So the expression becomes:

$\frac{1}{(u-1)(v-1)} + \frac{1}{(u-1)(w-1)} + \frac{1}{(w-1)(v-1)}$ which on evaluating gives $\frac{u+v+w-3}{(u-1)(v-1)(w-1)} .............$ $(*)$

Also, $uv + vw + uw =\frac{1}{2} ((u+v+w)^2 - (u^2+v^2+w^2)) = -1006$

Expanding the denominator of $(*)$ ,& substituting values, we get the expression as $-\frac{2}{1007}$ , with $a=$ 2 .