Problem 2

Let

f(x)

and

g(x)

be given by

f(x)=\dfrac1x+\dfrac1{x-2}+\dfrac1{x-4}+\cdots+\dfrac1{x-2018}

and

g(x)=\dfrac1{x-1}+\dfrac1{x-3}+\dfrac1{x-5}+\cdots+\dfrac1{x-2017}

. Prove that

|f(x)-g(x)|>2

for any non-integer real number

x

satisfying

0<x<2018

Step 6 of 6: Bound each piece and conclude

Previous step Next step

\frac1{x-1}+\frac1{2-x}\ge 4,\qquad \frac3{x(x-3)}>-\frac32,\qquad d(x)>4-\frac32=\frac52>2

Detailed analysis

Since $(x-1)+(2-x)=1$ , the AM–HM inequality gives $\frac1{x-1}+\frac1{2-x}\ge\frac{4}{(x-1)+(2-x)}=4$ , with equality at $x=\tfrac32$ . For $1<x<2$ , the quadratic $x(x-3)$ ranges over $[-\tfrac94,-2)$ , so $|x(x-3)|>2$ and $\frac3{x(x-3)}>-\tfrac32$ . Adding these bounds gives $d(x)>4-\tfrac32=\tfrac52>2$ for $1<x<2$ , and by steps 3–4 this proves $d(x)=g(x)-f(x)>2$ on every interval $2n-1<x<2n$ ; by step 2 the same bound (applied with $f,g$ swapped) covers $2n<x<2n+1$ , so $|f(x)-g(x)|>2$ for all non-integer $x$ with $0<x<2018$ .