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Let $x,y,z$ be positive real numbers such that $x^3+y^3+z^3=3$. Find the minimum value of $$P=\frac{1}{2-x}+\frac{1}{2-y}+\frac{1}{2-z}.$$

I think that we need to show that $\dfrac{1}{2-x} \ge f(x)$, where $0<x<\sqrt[3]{3}$ but I still have no solution.

I really appreciate if some one can help me. Thanks!

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    $\begingroup$ hint: Lagrange mulitplier surely works. $\endgroup$ – DeepSea Aug 26 '15 at 4:44
  • $\begingroup$ Can you post it? $\endgroup$ – kimtahe6 Aug 26 '15 at 4:50
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I make a sketch of the work here by Lagrange Multiplier. I am sure there are other more "elegant" techniques from exotic inequalities like AM-GM, Schur, CS, etc...I would be more than happy to see a proof of such work.Meanwhile, we can take this path... We are led to: $(2-x)x=(2-y)y=(2-z)z = \sqrt{3\lambda}\to \text { WLOG } x = y, x+z = 2$ is the worse case since the ideal case is $x = y = z = 1$. Thus $f(x,y,z) = f(x)=\dfrac{2}{2-x}+\dfrac{1}{x}$. From this you can use derivative to finish and find the minimum value of $f(x)$.

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  • $\begingroup$ Thank you very much Ganymede. I'll try it from your hints. $\endgroup$ – kimtahe6 Aug 26 '15 at 5:11
  • $\begingroup$ I am sure that the minimum value of P less than 3. I am calculating since your hints. $\endgroup$ – kimtahe6 Aug 26 '15 at 5:19
  • $\begingroup$ I define the Lagrange to be $L(x, y, z, t)=\dfrac{1}{2-x}+\dfrac{1}{2-y}+\dfrac{1}{2-z}-t(x^3+y^3+z^3-3)$. The critical points are solutions to $L_x=0,L_y=0,L_z=0, L_t=0$, then $(2-x)x=(2-y)y=(2-z)z = \dfrac{1}{\sqrt{3t}}=\sqrt{3 \lambda}$. I found $x=2(\sqrt{2}-1)$. Is it right? $\endgroup$ – kimtahe6 Aug 26 '15 at 6:48
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    $\begingroup$ Its true. That's what I got also. $\endgroup$ – DeepSea Aug 26 '15 at 6:53
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    $\begingroup$ All cases lead to those 2 cases. $\endgroup$ – DeepSea Aug 26 '15 at 7:04

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