Let $a, b, c, d \in R^+$ such that $a + b + c + d = 1$. Prove that $\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq \frac{1}{8}$ Let $a, b, c, d \in R^+$ such that $a + b + c + d = 1$. Prove that,
$$\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq \frac{1}{8}$$
Well from their sum we do get that $\frac{1}{4} \geq \sqrt[4]{abcd}$
$$\Rightarrow \frac{1}{4^8} \geq a^2b^2c^2d^2$$
and applying AM-GM on LHS of the given inequality we get ,
$$\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq 4 \cdot \sqrt[4]{\frac{a^3b^3c^3d^3}{(a+b)(b+c)(c+d)(d+a)}}$$
and  $(a+b)(b+c)(c+d)(d+a) \geq 16 \cdot abcd$ or
$$\frac{a^3b^3c^3d^3}{(a+b)(b+c)(c+d)(d+a)} \leq 16 \cdot a^2b^2c^2d^2 \leq 4^2 \cdot \frac{1}{4^8}= \frac{1}{4^6}$$
$$\Rightarrow \frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq 4 \cdot\frac{1}{8}=\frac{1}{2} > \frac{1}{8} \blacksquare.$$
Is this proof correct ? Did i miss any details? My doubt really stems from the fact that i didnt get $\frac{1}{8}$ directly but $\frac{1}{2}$, which makes my resultant inequality strict instead of being $\geq$ and it makes me wonder whether my proof is right. Thanks.
EDIT: Well guys i haven't read Titu's Lemma or Holder's inequality just yet though both of them do seem very powerful. I guess i'll just come to this question later when m done with those topics. Thanks for your help. Also I was just wondering whether it is possible to do it purely using AM-GM or maybe WAM-WGM ? Thanks again.

No. Your proof is not correct. 
After your first step we need to prove a wrong for $a\rightarrow0^+$ inequality. 
I think it's better to use Holder: $$\sum_{cyc}\frac{a^3}{b+c}\geq\frac{(a+b+c+d)^3}{4\sum\limits_{cyc}(b+c)}=\frac{1}{8}.$$
