<s>[INST] prove $\sum\limits_{cyc} \frac {a^3} {b+c+d} \geq \frac {1} {3}$ 
Show that if $a$, $b$, $c$ and $d$ are non-negatives  and $ab+bc+cd+da=1$ then:
  $$\sum\limits_{cyc}  \frac {a^3} {b+c+d} \geq \frac {1} {3}$$

yet again it should be solved with Cauchy inequality.
thing i have done so far:
$(\sum\limits_{cyc}  \frac {a^3} {b+c+d})\times(\sum\limits_{cyc} a(b+c+d)) \geq (\sum\limits_{cyc}  a^2)^2$
so my problem is simplified to proving this:
$\frac {(\sum\limits_{cyc}  a^2)^2} {(\sum\limits_{cyc} a(b+c+d))} \geq \frac {1} {3}$
$3 \times (\sum\limits_{cyc}  a^2)^2 \geq \sum\limits_{cyc} a(b+c+d)$
$3 \times (a^2+b^2+c^2+d^2)^2 \geq 2(ab+ac+ad+bc+bd+cd)$
someone said to me if i play around with AM-GM it could be solved and i'm almost there
my idea is this right now:
prove $(a^2+b^2+c^2+d^2) \geq ab+bc+cd+da=1$ proved(with help of Jineon Baek hint)
prove $3(a^2+b^2+c^2+d^2) \geq 2(ab+ac+ad+bc+bd+cd)$ proved(with help of Jineon Baek hint)
 [/INST] To finish your idea, just add up all inequalities of type $a^2+b^2 \geq 2ab$.
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