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.

\begin{align*}
\sum \frac{a^3}{b+c+d} &\geq \frac{4(a^3+b^3+c^3+d^3)}{3(a+b+c+d)}\\
&\geq \frac{(a+c)^3+(b+d)^3}{3(a+b+c+d)}\\
&=\frac{1}{3}((a+c)^2+(b+d)^2-(a+c)(b+d))\\
&\geq \frac{(a+c)(b+d)}{3}\\
&=\frac{1}{3}
\end{align*}
