For $0<r<1$, consider a function $f:[0,r)\to \mathbb{R}$ defined as $$f(x)=\dfrac{r-x}{1-xr}.$$ Then I want to prove $|f(x)|\leq r$. How I attempted to prove this is: $|f(x)|\leq \dfrac{r+r}{1-r^2}$ (using triangle inequality and the inequality $|a-b|\geq ||a|-|b||$ ). But for the expression on left to be less than r, one need $\dfrac{2}{1-r^2}\leq 1$ (which I am unable to see). What other approach I can use to prove it or am I missing something obvious in these steps.
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2Square it. You get an equivalent quadratic inequality. – MathematicsStudent1122 Dec 12 '21 at 20:26
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1Alt. hint (for $,r \ne 0,$): write it as $,\left|\dfrac{1-\frac{x}{r}}{1-xr}\right| \leq 1,$, and use that $,0 \lt xr \lt \frac{x}{r} \le 1,$. – dxiv Dec 12 '21 at 21:39
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Note that for the given conditions $f(x)\ge 0$ so we can get rid of the absolute value:
- for $x\in[0,r]$ you have $r-x\ge 0$
- also $1-xr\ge 1-r^2>0$ for $r\in(0,1)$
We can then study the sign of $\ r-f(x)=r-\dfrac{r-x}{1-xr}=\dfrac{(1-r^2)x}{1-xr}\ge 0$ for the same reasons as above.
zwim
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$$ f(x)=\frac{r-x}{1-xr} = \frac 1r + \frac{r^2-1}{r(1-xr)} $$ is decreasing on $[0, r]$, so that $$ r = f(0) \ge f(x) \ge f(r) = 0 $$ on that interval.
Martin R
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