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I would appreciate if somebody could help me with the following problem:

Q: find $x$

$$\frac{1}{x-1}+ \frac{2}{x-2}+ \frac{3}{x-3}+\cdots+\frac{10}{x-10}\geq\frac{1}{2} $$

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3 Answers 3

If the left side is $$f(x)=\sum_{k=1}^{10} \frac{k}{x-k},$$ then the graph of $f$ shows that $f(x)<0$ on $(-\infty,1)$, so no solutions there. For each $k=1..9$ there is a vertical asymptote at $x=k$ with the value of $f(x)$ coming down from $+\infty$ immediately to the right of $x=k$ and crossing the line $y=1/2$ between $k$ and $k+1$, afterwards remaining less than $1/2$ in the interval $(k,k+1)$. This gives nine intervals of the form $(k,k+a_k]$ where $f(x) \ge 1/2$, and there is a tenth interval in which $f(x) \ge 1/2$ beginning at $x=10$ of the form $(10,a_{10}]$ where $a_{10}$ lies somewhere in the interval $[117.0538,117.0539].$ The values of the $a_k$ for $k=1..9$ are all less than $1$, starting out small and increasing with $k$, some approximations being $$a_1=0.078,\ a_2=0.143,\ a_3=0.201,\ ...\ a_9=0.615.$$ The formula for finding the exact values of the $a_k$ for $k=1..10$ is a tenth degree polynomial equation which maple12 could not solve exactly, hence the numerical solutions above.

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the 10th degree polynomial is irreducible, by the way. –  Ewan Delanoy Mar 13 '13 at 16:28
    
Thanks. I didn't check that, but assumed it since maple didn't even give a smaller "RootOf" than degree ten. –  coffeemath Mar 14 '13 at 13:14

Let $x$ be any number such that $10 \lt x \lt 21$. Then for each $k\in [1,10]$ we have $21k \gt x \gt k$, hence $20k \gt x-k \gt 0$, so $\frac{k}{x-k} \gt \frac{1}{20}$. Summing from $k=1$ to $10$, we obtain the desired inequality.

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There are other solutions near $k$ for $k=1...9$ and beyond 10 the solution set extends to about 117. –  coffeemath Mar 13 '13 at 12:46

this picture is show what coffeemath said: enter image description here

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