<s>[INST] Let $x = (a^2+a+1)^{1/2}-(a^2-a+1)^{1/2}$ and $a$ belongs to real number find all possible value of $x$. Let $A = (-1/2, \sqrt{3}/2)$, $B = (1/2, \sqrt{3}/2)$, and $P = (a, 0)$. Then $P$ is a point on the $X$-axis and we are looking for all possible values of $d = PA - PB$.
By the Triangle Inequality, $PA - PB < AB = 1$. And it is clear
that all the values $-1 < d < 1$ are indeed obtainable. In fact, for such
a $d$, a half hyperbola of all points $Q$ such that $QA - QB = d$ is well
defined. (Points $A$ and $B$ are foci of the hyperbola.) Since line $AB$ is parallel to the $X$-axis, this half hyperbola intersects the
$X$- axis, i.e., $P$ is well defined.
I know that at point $A$ and $B$ the two above $\sqrt{\cdot}$ function will obtain minimum value respectively
I am not able to understand why are we looking for all the possible value of $d$, what is the relationship of $d$ with the possible value of $x$?
This is question number 14 from 101 problems in algebra (USAIMO)
 [/INST] Since
$a^{1/2}-b^{1/2}
=(a^{1/2}-b^{1/2})\dfrac{a^{1/2}+b^{1/2}}{a^{1/2}+b^{1/2}}
=\dfrac{a-b}{a^{1/2}+b^{1/2}}
$,
$d(a)
=(a^2+a+1)^{1/2}-(a^2-a+1)^{1/2}
=\dfrac{2a}{(a^2+a+1)^{1/2}+(a^2-a+1)^{1/2}}
$.
Also,
by the generalized mean inequality,
$\sqrt{uv}
\le (\dfrac{\sqrt{u}+\sqrt{v}}{2})^2
\le \dfrac{u+v}{2}
$
or
$2\sqrt[4]{uv}
\le\sqrt{u}+\sqrt{v}
\le \sqrt{2(u+v)}
$
so
$d(a)
\ge \dfrac{2a}{\sqrt{4(a^2+1)}}
= \dfrac{a}{\sqrt{a^2+1}}
$
and
$\begin{array}\\
d(a)
&\le \dfrac{2a}{2\sqrt[4]{(a^2+a+1)(a^2-a+1)}}\\
&= \dfrac{a}{\sqrt[4]{(a^2+1)^2-a^2}}\\
&= \dfrac{a}{\sqrt[4]{a^4+a^2+1}}\\
&\le 1\\
\end{array}
$
So the range of values is from
$0$ to $1$.
</s>