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let $a,b,c>0$,and $$abc=1$$ show that $$b(a-1)(c-1)+c(b-1)(a-1)+a(c-1)(b-1)\le 0$$

since $$b(a-1)(c-1)=b(ac-a-c+1)=abc-ab-bc+b=1-ab-bc+b$$ so we only prove $$3-2(ab+bc+ac)+a+b+c\le 0 $$

oh,this inequality is wrong.let $a=0.1,b=0.1,c=100$,then we have

http://www.wolframalpha.com/input/?i=0.1%280.1-1%29%28100-1%29%2B100%280.1-1%29%280.1-1%29%2B0.1%2899%29*%280.1-1%29

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    $\begingroup$ For a=9,b=1/3,c=1/3, I find that your expression is 4/9, hence not $\leq 0$. Where is my error ? $\endgroup$ – Kelenner Aug 16 '14 at 19:15
  • $\begingroup$ ya, a=4,b=c=1/2, LHS <0. so this inequality is no sense at all. $\endgroup$ – chenbai Aug 17 '14 at 8:59
  • $\begingroup$ This question appears to be off-topic because it is incorrect. $\endgroup$ – Macavity Aug 17 '14 at 16:49
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prove:$$3-2\sum_{cyc}ab+\sum_{cyc}a\leq0$$Proof:

Let:$$a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$$and:$$x\geq y\geq z,x=y+m,z=y-t;m,t\geq 0$$then you need prove:$$3-2\sum_{cyc}\frac{x}{z}+\sum_{cyc}\frac{z}{x}\leq 0$$noticed:$$\sum_{cyc}\frac{x}{z}\geq 3$$you just need prove:$$\sum_{cyc}\frac{x}{z}\geq \sum_{cyc}\frac{z}{x}$$$$\Longleftrightarrow x^2(y-z)+y^2(z-x)+z^2(x-y)\geq 0$$ $$\Longleftrightarrow mt(m+t)\geq 0$$and this is ture.

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