I want to simplify the expression \begin{equation} A = \frac{\sqrt{1-\sqrt{1-x^2}}\Big[\sqrt{(1+x)^3} + \sqrt{(1-x)^3} \Big]}{2-\sqrt{1-\sqrt{1-x^2}}} \end{equation} and find the minimum value of $\displaystyle \sqrt{2}A+\frac{1}{x}$.

I can only simplify to this point \begin{equation} A = \frac{2x\Big( 2-\sqrt{1-x^2} \Big)}{2\sqrt{2}-\sqrt{1+x}+\sqrt{1-x}}. \end{equation} However, I am not sure if I can go any further. The next part of the question is to find the minimum value of $\displaystyle \sqrt{2}A+\frac{1}{x}$, which is obviously not so easy since I am not supposed to use differentiation to find optimal values.

  • $\begingroup$ if $x$ can be any value in(-1,1), then A have limitation, but 1/x don't have globe max or min. which means the min you want to get is only local min. Are you sure that is your question? $\endgroup$ – chenbai May 24 '15 at 14:10
  • $\begingroup$ This problem was given to me as it is, and I can only use pre-cal to solve. However, you're right. I looked at the graph, and it seemed we only have local min in $[-1,1]$. $\frac{1}{2}$ seems to be its local min. $\endgroup$ – dh16 May 24 '15 at 16:59
  • $\begingroup$ If we restrict to $0<x\leq 1$, we have a minimum value. $\endgroup$ – dh16 May 24 '15 at 17:18
  • $\begingroup$ the min will be got at $x=0.6$ roughly. I don't think precaculus can solve it. you may have to use numeric method. $\endgroup$ – chenbai May 25 '15 at 1:17

Try this: $u = \sqrt{1+x}, v = \sqrt{1-x} \to \sqrt{1-x^2} = uv, 2 = u^2+v^2$, and consider $f(u,v) = ...$

  • $\begingroup$ I let $x=\sin 2\theta$ and get to $\displaystyle \frac{2\sin 2\theta(2-|\cos 2\theta|)}{2\sqrt{2}-|\sin\theta + \cos\theta| + |\sin\theta - \cos\theta|}$. Then what should I do next? Thank you. $\endgroup$ – dh16 May 24 '15 at 4:49
  • $\begingroup$ What is the mistake? I checked, but couldn't see it. If I assume the interval for $\theta$, I can get rid of the abs value. $\endgroup$ – dh16 May 24 '15 at 5:01
  • $\begingroup$ I tried it, but it did not simplify any further. This problem is hard. $\endgroup$ – dh16 May 24 '15 at 16:53
  • $\begingroup$ I think you need to do it in two steps, ...i will get back on this a few hours.... $\endgroup$ – DeepSea May 24 '15 at 16:57

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