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.
 A: Credits to Calvin Lin for the hint to apply Titu's Lemma, but I don't find the remaining part of the proof obvious.
Applying Titu's lemma yields:
\begin{align*}
\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} &= \frac{a^4}{ab+ac}+\frac{b^4}{bc+bd}+\frac{c^4}{cd+ca}+\frac{d^4}{da+db} \\
&\geq \frac{(a^2 + b^2 + c^2 + d^2)^2}{ab + ac + bc + bd + cd + ca + da + db}
\end{align*}
Applying Cauchy-Schwarz Inequality to the denominator yields:
\begin{align*}
ab + ac + bc + bd + cd + ca + da + db &\leq \sqrt{\left(2a^2 + 2b^2 + 2c^2 + 2d^2 \right)^2}  \\
&= 2(a^2 + b^2 + c^2 + d^2)
\end{align*}
Therefore:
\begin{align*}
\frac{(a^2 + b^2 + c^2 + d^2)^2}{ab + ac + bc + bd + cd + ca + da + db} \geq \frac{1}{2}(a^2 + b^2 + c^2 +d^2)
\end{align*}
Applying Cauchy-Schwarz inequality again with $(a^2,b^2,c^2,d^2) \cdot (1,1,1,1)$ yields::
\begin{align*}
4(a^2 + b^2 + c^2 + d^2) \geq (a + b + c + d)^2 = 1
\end{align*}
Therefore:
\begin{align*}
\frac{1}{2}(a^2 + b^2 + c^2 +d^2) \geq \frac{1}{2}\left(\frac{1}{4}\right) = \frac{1}{8}
\end{align*}
A: 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: $$\sum_{cyc}\frac{a^3}{b+c}\geq\frac{(a+b+c+d)^3}{4\sum\limits_{cyc}(b+c)}=\frac{1}{8}.$$
