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Algebra Difficulty 5.9 AIME, harder Prove it

Show that if x0>x1>⋯>xnx_{0}>x_{1}>\cdots>x_{n}, then

x0+1x0−x1+1x1−x2+⋯+1xn−1−xn⩾xn+2n x_{0}+\frac{1}{x_{0}-x_{1}}+\frac{1}{x_{1}-x_{2}}+\cdots+\frac{1}{x_{n-1}-x_{n}} \geqslant x_{n}+2 n

Solution

The trick is to make the terms x0−x1,x1−x2,…,xn−1−xnx_{0}-x_{1}, x_{1}-x_{2}, \ldots, x_{n-1}-x_{n} appear thanks to a telescoping on the term x0−xnx_{0}-x_{n}. Indeed, we have:

x0−xn=(x0−x1)+(x1−x2)+⋯+(xn−1−xn) x_{0}-x_{n}=\left(x_{0}-x_{1}\right)+\left(x_{1}-x_{2}\right)+\cdots+\left(x_{n-1}-x_{n}\right)

so that the required inequality is equivalent to

(x0−x1)+(x1−x2)+⋯+(xn−1−xn)+1x0−x1+1x1−x2+⋯+1xn−1−xn⩾2n \left(x_{0}-x_{1}\right)+\left(x_{1}-x_{2}\right)+\cdots+\left(x_{n-1}-x_{n}\right)+\frac{1}{x_{0}-x_{1}}+\frac{1}{x_{1}-x_{2}}+\cdots+\frac{1}{x_{n-1}-x_{n}} \geqslant 2 n

or, by pairing each term with its reciprocal, to

(x0−x1+1x0−x1)+(x1−x2+1x1−x2)+⋯+(xn−1−xn+1xn−1−xn)⩾2n \left(x_{0}-x_{1}+\frac{1}{x_{0}-x_{1}}\right)+\left(x_{1}-x_{2}+\frac{1}{x_{1}-x_{2}}\right)+\cdots+\left(x_{n-1}-x_{n}+\frac{1}{x_{n-1}-x_{n}}\right) \geqslant 2 n

which is true by applying the result of Exercise 1 for x=x0−x1,x1−x2,…,xn−1−xnx=x_{0}-x_{1}, x_{1}-x_{2}, \ldots, x_{n-1}-x_{n}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.