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If $A > 0, |x| < \sqrt{A}$ prove that $\frac{1}{2}\left|x + A\frac{1}{x}\right| > \sqrt{A}$

My work:

\begin{align*} |x| <& \sqrt{A} \\ \frac{1}{|x|} >& \frac{1}{\sqrt{A}} \\ A\frac{1}{|x|} >& \sqrt{A} \\ \end{align*}

\begin{align*} \frac{1}{2}\left|x + A\frac{1}{x}\right| &\le \frac{1}{2} \left(|x| + A \frac{1}{|x|}\right)\\ \end{align*}

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    $\begingroup$ Do you know the $AM-GM$ inequality? $\endgroup$ – AnotherJohnDoe Oct 3 '17 at 18:55
  • $\begingroup$ No, but reading it now. $\endgroup$ – clay Oct 3 '17 at 18:58
  • $\begingroup$ For $a\ge 0$, and $b\ge 0$, we have $(\sqrt a - \sqrt b)^2\ge0\;\;\Rightarrow a+b\ge 2\sqrt{ab}$ $\endgroup$ – AnotherJohnDoe Oct 3 '17 at 19:02
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squaring the given inequality we get $$\frac{1}{4}\left(x^2+\frac{A^2}{x^2}+2A\right)\geq 4A$$and this is equivalent to $$\left(x-\frac{A}{x}\right)^2\geq 0$$ since $$|x|<\sqrt{A}$$ it must be $$A>0$$

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