231. A conditioned minimum value.

by Virgil Nicula, Feb 25, 2011, 4:02 PM

Proposed problem. For $a>0$ and $b>0$ ascertain the minimum value of $y>a$ for which is truly the implication $\boxed{\ x\ne 0\ \wedge\ |x-y|\le a\ \implies\ \left|\frac 1x-\frac 1y\right|\le \frac 1b\ }$ .

Proof. Denote $A(y)=\left|x\in\mathbb R^*\right|\left|\ |x-y|\le a\ \right|$ and $B(y)=\left|x\in\mathbb R^*\right|\ \left|\frac 1x-\frac 1y\le \frac 1b\ \right|$ . Thus, $\left\|\begin{array}{ccc} x\in A(y) & \iff & \frac {1}{y+a}\le\frac 1x\le\frac {1}{y-a}\\\\ x\in B(y) & \iff & \frac 1y-\frac 1b\le\frac 1x\le\frac 1y+\frac 1b\end{array}\right\|$ .

Since $A(y)\subset B(y)$ obtain that $\frac 1y-\frac 1b\le\frac {1}{y+a}$ and $\frac {1}{y-a}\le\frac 1y+\frac 1b$ , i.e. $\max\left\{\frac 1y-\frac {1}{y+a}\ ,\ \frac {1}{y-a}-\frac 1y\right\}\le\frac 1b$ , i.e. $\frac {a}{y(y-a)}\le \frac 1b$ . Hence

$f(y)\equiv y^2-ay-ab\ge 0$ . Observe that $f(a)=-ab$ and $f(a+b)=b^2>0$ , what means that the equation $f(y)=0$ has two real roots $y_1$ , $y_2$

and $y_1<a<y_2<a+b$ . Therefore, $\min\{a<y|A(y)\subset B(y)\}=$ $\min\{a<y|f(y)\ge 0\}=y_2=$ $\frac 12\cdot\left(a+\sqrt {a^2+4ab}\right)$ is required minimum value.

Remark. For $b\le 2a$ have $y_2\in\left[a+\frac b2,a+b\right)$ , where the inclusion $A(y_2)\subset B(y_2)$ is strictly because $\frac {1}{y_2}-\frac 1b<\frac {1}{y_2+a}$ and $\frac {1}{y_2+a}=\frac {1}{y_2}+\frac 1b$ .
This post has been edited 8 times. Last edited by Virgil Nicula, Nov 22, 2015, 2:42 PM

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