How prove this inequality $\sum\limits_{cyc}\frac{a^3}{b+c+d}\ge \dfrac{1}{3}$,if $\sum\limits_{cyc}a\sqrt{bc}\ge 1$ 
Let $a$, $b$, $c$ and $d$ be non-negative numbers such that
  $$a\sqrt{bc}+b\sqrt{cd}+c\sqrt{da}+d\sqrt{ab}\ge 1.$$
  Show that
  $$\dfrac{a^3}{b+c+d}+\dfrac{b^3}{a+c+d}+\dfrac{c^3}{a+b+d}+\dfrac{d^3}{a+b+c}\ge \dfrac{1}{3}$$ 

My try:use Cauchy-Schwarz inequality we have
 $$\sum_{cyc}\dfrac{a^3}{b+c+d}\left(\sum_{cyc}(b+c+d)\right)\ge \left(\sum_{cyc}a^{3/2}\right)^2$$
then I can't contious, and I found this simaler problem
IMO 1990 problem:Let $a,b,c,d\ge 0$,and such 
$$ab+bc+cd+ad= 1$$
show that
$$\dfrac{a^3}{b+c+d}+\dfrac{b^3}{a+c+d}+\dfrac{c^3}{a+b+d}+\dfrac{d^3}{a+b+c}\ge \dfrac{1}{3}$$ 
solution can see 
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=360441&sid=a8e8d6ea18dc9e7121bccaedbd15f41f#p360441
 A: WLOG, to simplify expressions, we can consider the squares of $a, b, c, d$ instead and let $s=a^2+b^2+c^2+d^2$,  so the problem can be rephrased to show that for $a, b, c, d \ge 0$,
$$\color{blue}{\sum_{cyc} a^2bc\ge 1 \implies \sum_{cyc} \frac{a^6}{s-a^2} \geqslant \frac13}$$ 
Convexity of $t \mapsto \dfrac{t^3}{s-t}$ and Jensen's inequality gives us:
$$\sum_{cyc} \frac{a^6}{s-a^2} \ge 4\frac{\left(\frac{s}4\right)^3}{s-\frac{s}4} = \frac{s^2}{12}$$
So it is enough to show that $\displaystyle \sum_{cyc} a^2bc \ge 1 \implies s^2 \ge 4$
We have
$$1\le \sum_{cyc} a^2bc  = ac(ab+cd)+bd(ad+bc)$$
If $ad+bc \le ab+cd$, then we get from the above,
\begin{align}
1 &\le (ac+bd)(ab+cd) \\
&\color{red}{\le}  \left(\frac{ac+bd+ab+cd}2\right)^2 =\left(\frac{(a+d)(b+c)}2\right)^2 \\
&\color{red}{\le} 4 \left(\frac{a+b+c+d}4\right)^4 \\
&\color{green}{\le} 4 \left(\frac{a^2+b^2+c^2+d^2}4\right)^2 = \frac{s^2}4 \\
\end{align}
where we have used $\color{red}{AM-GM}$ and $\color{green}{AM-QM}$ inequalities.
OTOH, if $ab+cd \le ad+bc$, proceeding along similar lines, by symmetry we get the same result. Hence proved.
A: By Holder $$\sum_{cyc}\frac{a^3}{b+c+d}\geq\frac{(a+b+c+d)^3}{4\sum\limits_{cyc}(b+c+d)}=\frac{(a+b+c+d)^2}{12}.$$
Thus, it remains to prove that
$$(a+b+c+d)^2\geq4(a\sqrt{bc}+b\sqrt{cd}+c\sqrt{da}+d\sqrt{ab}).$$
Let $\{a,b,c,d\}=\{x,y,z,t\}$, where $x\geq y\geq z\geq t$.
Hence, by Rearrangement, AM-GM and C-S we obtain:
$$4(a\sqrt{bc}+b\sqrt{cd}+c\sqrt{da}+d\sqrt{ab})=4(\sqrt{a}\sqrt{abc}+\sqrt{b}\sqrt{bcd}+\sqrt{c}\sqrt{cda}+\sqrt{d}\sqrt{dab})\leq$$
$$\leq4(\sqrt{x}\sqrt{xyz}+\sqrt{y}\sqrt{xyt}+\sqrt{z}\sqrt{xzt}+\sqrt{t}\sqrt{yzt})=4\left(\sqrt{xz}+\sqrt{yt}\right)\left(\sqrt{xy}+\sqrt{zt}\right)\leq$$
$$\leq4\left(\frac{\sqrt{xz}+\sqrt{yt}+\sqrt{xy}+\sqrt{zt}}{2}\right)^2=\left((\sqrt{x}+\sqrt{t})(\sqrt{y}+\sqrt{z})\right)^2\leq$$
$$\leq\left(\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{t}}{2}\right)^4\leq\frac{1}{16}\left(\sqrt{4(x+y+z+t)}\right)^4=$$
$$=(x+y+z+t)^2=(a+b+c+d)^2.$$
Done!
