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I have been trying to prove without success the following inequality. If $a,b,c,d>0$ and $a^2+b^2+c^2+d^2=1$ then prove that $a+b+c+d+\frac{1}{abcd}\geq 18$.

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You should describe the approaches you did try, even if unsuccessful. –  Casteels Feb 27 '14 at 9:57
This is just a guess and I have not worked it out yet... Put $a = \cos\phi\cos\theta$, $b = \sin\phi\cos\theta$, $c = \cos\phi\sin\theta$ and $d = \sin\phi\sin\theta$ with $\theta,\phi\in[0,\pi/2]$. Does that help? –  L__ Feb 27 '14 at 10:06
@L__: be careful, you only parametrized the intersection of the hypersphere with $ad=bc$... –  zozoens Feb 27 '14 at 10:12

1 Answer 1

up vote 2 down vote accepted

First, when $a=b=c=d=\dfrac{1}{2}$, the equality holds.

Now try to use $$ \sum a+32\times\frac{1}{32abcd}\ge 36 \sqrt[36]{\frac{abcd}{(32abcd)^{32}}}. $$

Notice also that $1=a^2+b^2+c^2+d^2\ge 4\sqrt{abcd}$, we complete the proof.

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Ma Ming, please explain in details how did you get right hand side of the inequality. In particulat how did you get 36? –  user61810 Mar 2 '14 at 5:56
@user61810 $4+32=36$. –  Ma Ming Mar 2 '14 at 9:07
Write down all details please. You have skipped something which is not obvious at all. –  user61810 Mar 2 '14 at 14:50

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