<s>[INST] 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.
 [/INST] Just another way, we can use a lesser used but quite useful generalisation of the Radon inequality (see here for e.g.).  Reproducing for ease of reference, for positive reals $a_i, b_i, r,s$ s.t. $r\geqslant s+1$ it says:
$$\sum_{i=1}^n \frac{a_i^r}{b_i^s} \geqslant \frac{\left(\sum_i a_i \right)^r}{n^{r-s-1}\;\left( \sum_i b_i\right)^s}$$
It can now be easily noted that for our problem$$\sum \frac{a^3}{b+c+d}\geqslant \frac{(a+b+c+d)^3}{4\cdot3\cdot(a+b+c+d)}=\frac1{12}(a+b+c+d)^2$$
Now by AM-GM, $(a+b+c+d)^2\geqslant 4(a+c)(b+d)=4$ finishes this.
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