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Let $a,b,c>0 ,\sum\limits_{cyc} a^{3}=3$. How prove $\sum \frac{ab}{c} \geq 3.$
My try:
$ a^3+b^3+c^3=3$

$a^2b^2+a^2c^2+b^2c^2 \ge 3abc$ so

$ (a+b+c)^3-3(a+b+c)(ab+ac+bc)+3abc=3$

$(ab+ac+bc)^2-2(a+b+c)abc \ge 3abc$

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    $\begingroup$ What are you summing over? $\endgroup$ – Jendrik Stelzner Jan 2 '16 at 13:03
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    $\begingroup$ It is not clear what does it means by $\sum a^3$ and $\sum ab/c$ $\endgroup$ – sinbadh Jan 2 '16 at 13:03
  • $\begingroup$ Are you trying to show $a^2b^2+b^2c^2+c^2a^2 \ge 3abc$ given $a^3+b^3+c^3 = 3$ ? $\endgroup$ – Henry Jan 2 '16 at 13:04
  • $\begingroup$ I try that:$$x=a+b+c$$, $$y=ab+ac+bc$$, $$z=abc$$ so $$x^3-3xy+3z=3$$ and $$y^2-2xz \ge 3z$$ $\endgroup$ – piteer Jan 2 '16 at 13:11
  • $\begingroup$ @Henry yes am trying that $\endgroup$ – piteer Jan 2 '16 at 13:21
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The inequality is false: take $a=\frac{1}{\sqrt[3] 6}$,$b=\frac{1}{\sqrt[3] 6}$,$c=\frac{\sqrt[3] 8}{\sqrt[3] 3}$, you obtain

$$ \sim 2.9918... \ge 3.$$

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  • $\begingroup$ ya, the inequality is only true when $a+b+c=3$ $\endgroup$ – chenbai Jan 3 '16 at 7:25

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