Prove for a,b,c,d > 0 $\frac{a^3}{b+c+d} + \frac{b^3}{a+c+d} + \frac{c^3}{a+b+d} + \frac{d^3}{a+b+c} \ge \frac{1}{3}$ where $ab + bc + cd + da = 1$ The task requires for Cauchy-Schwarz inequality to be used.
My attempt: Using Titu's Lemma (direct consequence of the inequality), I got:
$\frac{a^4}{a(b+c+d)} + \frac{b^4}{b(a+c+d)} + \frac{c^4}{c(a+b+d)} + \frac{d^4}{d(a+b+c)} \ge \frac{(a^2 + b^2 + c^2 + d^2)^2}{a(b+c+d) \ + \ b(a+c+d) \ + \ c(a+b+d) \ + \ d(a+b+c)}$
Using the condition that $ab + bc + cd + da = 1$, the RHS results in $\frac{(a^2+b^2+c^2+d^2)^2}{2+2(ac+bd)}$.
Nothing more seems to help. Opening up the numerator or using AM-GM on it doesn't work and I have no idea on how to turn this result in a $\frac{1}{3}$ fraction.

By Cauchy and AM-GM inequality,
$$\sum\frac{a^3}{b+c+d}\ge\frac{\left(\sum a^2\right)^2}{2\sum ab}\ge\frac{\left(\sum a^2\right)^2}{3\sum a^2}=\frac{\sum a^2}3\ge\frac13.$$
The last "$\ge$" is equivalent to $\sum a^2\ge1$. By rearrangement inequality
$$a^2+b^2+c^2+d^2\ge ab+bc+cd+da=1.$$We're done.
