Let we think about 2-d orthogonal coordinate $xOy$. set point $A(a,a)$, where $a \in \Bbb{R}$. and $P$ is a point in the function $$y=\frac{1}{x}$$. if $$|PA|\geq 2\sqrt{2}$$. find all $a\in \Bbb{R}$ that satisfied this inequality.


My approach:

Assume $P(t,\frac{1}{t})$, then $|PA|=\sqrt{(t-a)^2+(\frac{1}{t}-a)^2}$. so I need to find $a$, such that $$(t-a)^2+(\frac{1}{t}-a)^2 \geq 8$$ for all $t >0$. then I got an inequality for $a$ and $t$. and I try to treat this as an equation of $a$. then solve it, we have root $$t+\frac{1}{t}+\sqrt{18-t^2-\frac{1}{t^2}}$$ and $$t+\frac{1}{t}-\sqrt{18-t^2-\frac{1}{t^2}}$$. If for all $t>0$ ,the above inequality hold ,equal to $$a \geq t+\frac{1}{t}+\sqrt{18-t^2-\frac{1}{t^2}}$$ or $$a \leq t+\frac{1}{t}-\sqrt{18-t^2-\frac{1}{t^2}}$$

But I can't go any further. Please help me :(

  • $\begingroup$ Put $t+\frac{1}{t}=z$ and solve the quadratic in $z$.Now, the discriminant(which is a function of $a$) must be greater than equal to $0$ . Solve to get the answer. Also note that $|z|\geq2$. Solve the inequality and the answer will be the intersection of the values obtained from these two inequalities $\endgroup$ – idpd15 Apr 5 '14 at 10:37
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    $\begingroup$ $|PA|\geqslant2\sqrt2$ for every $P$ on the hyperbola $xy=1$ if and only if $|a|\geqslant\sqrt{10}$. $\endgroup$ – Did Apr 5 '14 at 15:06

You want to find $a$ s.t. $(t-a)^2+(\frac{1}{t}-a)^2 \geq 8$ is true for all non-zero $t \in \mathbb{R}$.

$$\iff \left(t^2 + \frac1{t^2} \right)-2a\left(t+\frac1t \right)+2a^2 \geqslant 8$$

Let $t+\frac1t = x$. Using this, we need for all $ \lvert x \rvert \ge 2$, the following inequality to hold. $$ (x-a)^2 \geqslant 10-a^2 \tag{1}$$ Clearly $a^2\geqslant 10$ ensures the RHS is negative and this inequality is true irrespective of $x$. As there may be other possible $a$, we investigate other cases.

Suppose $4 \leqslant a^2 < 10$. Then the LHS attains a minimum of $0$ while the RHS remains positive, so this doesn't work. Similarly, if $a^2 < 4$, then the minimum of the LHS is $\leqslant 4$ while the RHS is at least $6$, which means again the inequality does not hold for some $x$.

Thus the complete set of solutions is $a \in \mathbb{R} \setminus (-\sqrt{10}, \sqrt{10})$.

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