Prove $1 + \sum_{i=0}^n(\frac1{x_i}\prod_{j\neq i}(1+\frac1{x_j-x_i}))=\prod_{i=0}^n(1+\frac1{x_i})$ Prove the identity
$$1 + \sum_{i=0}^n \left(\frac1{x_i}\prod_{j\neq i} \left(1+\frac1{x_j-x_i} \right) \right)=\prod_{i=0}^n \left(1+\frac1{x_i} \right)$$
and hence deduce the inequality in Problem 2:
http://www.imc-math.org.uk/imc2010/imc2010-day2-solutions.pdf
Of course here we need all the $x_i$s distinct.
(By the way, this somehow resembles the Lagrange Interpolation formula.)
 A: Method 1
Let $f(z) = \prod_i (z - x_i)$, we have
$$\prod_{j\ne i}(1 + \frac{1}{x_j-x_i}) = \prod_{j\ne i}\frac{(x_i - 1) - x_j}{x_i - x_j}
= - \frac{f(x_i-1)}{f'(x_i)}$$
We can rewrite LHS of the equality as
$$1 - \sum_i \frac{f(x_i-1)}{x_if'(x_i)}
= 1 - \sum_i \frac{1}{2\pi i}\int_{C_i} \frac{f(z-1)}{zf'(x_i)(z-x_i)} dz
= 1 - \frac{1}{2\pi i}\int_{\sum_i C_i} \frac{f(z-1)}{zf(z)} dz
$$
where $C_i$ is a bunch of small circular contours surrounding $x_i$ counterclockwisely.
By deforming the contours, it is easy to see above contour integral is equal to
the difference of two contour integral, one with a big circle $C_{\infty}$ near infinity  and another small circle $C_o$ at origin:
$$\frac{1}{2\pi i}\int_{\sum_i C_i} \frac{f(z-1)}{zf(z)} dz
 = \frac{1}{2\pi i} \left( \int_{C_\infty} - \int_{C_o} \right) \frac{f(z-1)}{zf(z)} dz$$
Since $\displaystyle\;\;\frac{f(z-1)}{z f(z)} = \frac{1}{z} + O(\frac{1}{z^2})\;\;$ for large $z$, we have
$$\frac{1}{2\pi i} \int_{C_\infty} \frac{f(z-1)}{zf(z)} dz
 = \frac{1}{2\pi i} \int_{C_\infty} ( \frac{1}{z} + O(\frac{1}{z^2}) ) dz = 1$$
This cancels out the $1$ in LHS of inequality and we get
$$\text{LHS} = \frac{1}{2\pi i}\int_{C_o}\frac{f(z-1)}{zf(z)}dz = \frac{f(-1)}{f(0)}
=\frac{\prod_i(-1 - x_i)}{\prod_i ( -x_i )} = \prod_i (1+\frac{1}{x_i})
=\text{RHS}
$$
Method 2
If one don't want to use complex analysis, an alternate way is apply
Lagrange interpolation formula to polynomial $f(z-1) - f(z)$ whose degree is one less than the number of $x_i$, we have:
$$f(z-1) - f(z) 
= \sum_i \frac{f(z)(f(x_i-1) - f(x_i))}{(z-x_i)f'(x_i)}
= f(z) \sum_{i}\frac{f(x_i-1)}{(z-x_i)f'(x_i)}
$$
This implies 
$$1 + \sum_i \frac{f(x_i-1)}{(z-x_i)f'(x_i)} = \frac{f(z-1)}{f(z)}$$
Set $z = 0$, we obtain
$$\text{LHS} = 1 - \sum_i \frac{f(x_i-1)}{x_if'(x_i)} = \frac{f(-1)}{f(0)} = \text{RHS}.$$
