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If $a,b,c,d > 0$ and $abcd = 1$, prove that the following inequality holds:

$$\frac{1}{bc + cd + da -1} + \frac{1}{ab + cd + da -1} + \frac{1}{ab + bc + da -1} + \frac{1}{ab + bc + cd -1} \le 2$$


The problem I had is because it says that I need to prove that it's less than or equal to 2. With this I can't use any better known inequality (Cauchy-Schwarz, Holder, Minkowski...), because using that I'll prove that $\frac{1}{bc + cd + da -1} + \frac{1}{ab + cd + da -1} + \frac{1}{ab + bc + da -1} + \frac{1}{ab + bc + cd -1}$ is bigger than something, and even if I prove that that new inequality is true, that doesn't mean that the original inequallity holds.

Other my idea was to split the 2 in $4 \times \frac{1}{2}$ and i'll get:

$$\frac{1}{bc + cd + da -1} + \frac{1}{ab + cd + da -1} + \frac{1}{ab + bc + da -1} + \frac{1}{ab + bc + cd -1} \le \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}$$

Now I need to prove that:

$$\frac{1}{bc + cd + da -1} \le \frac{1}{2}$$ $$\frac{1}{ab + cd + da -1} \le \frac{1}{2}$$ $$\frac{1}{ab + bc + da -1} \le \frac{1}{2}$$ $$\frac{1}{ab + bc + cd -1} \le \frac{1}{2}$$

or:

$$bc + cd + da -1 \ge 2$$ $$ab + cd + da -1 \ge 2$$ $$ab + bc + da -1 \ge 2$$ $$ab + bc + cd -1 \ge 2$$

But if all inequalites hold then $a=b=c=d=1$ and that's as far I can go.

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    $\begingroup$ First use $\dfrac{1}{bc+da+cd -1} = \dfrac{1}{bc+da +cd - \sqrt{abcd}} \le \dfrac{2}{2cd + bc+da}$. By using $bc+da\ge 2\sqrt{abcd}$. Then use Cauchy Schwarz. $\endgroup$ – Yimin Apr 1 '13 at 18:34
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By inequality of arithmetic and geometric means, $bc+da\ge 2\sqrt{abcd}=2$, so $$\frac{1}{bc+cd+da-1}\le\frac{1}{1+cd}.\tag{1}$$ Similarly, $$\frac{1}{ab+bc+da-1}\le\frac{1}{1+ab}.\tag{2}$$ $(1)+(2)$ gives $$\frac{1}{bc+cd+da-1}+\frac{1}{ab+bc+da-1}\le\frac{1}{1+cd}+\frac{1}{1+ab}=\frac{2+ab+cd}{1+ab+cd+abcd}=1.\tag{3}$$ Similarly, $$\frac{1}{ab+cd+da-1}+\frac{1}{ab+bc+cd-1}\le 1.\tag{4}$$ The conclusion follows from $(3)+(4)$.

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  • $\begingroup$ Thank you man, I didn't think it'll be so simple :D $\endgroup$ – Stefan4024 Apr 1 '13 at 19:46
  • $\begingroup$ @Stefan4024: You are welcome! $\endgroup$ – 23rd Apr 1 '13 at 19:47

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