Using hypothesis $a=\frac{x}{y+z+t}$, $b=\frac{y}{z+t+x}$, $c=\frac{z}{t+x+y}$ and $d=\frac{t}{x+y+z}$, then $\sum \sqrt[3]{bcd}=\sum \sqrt[3]{\frac{yzt}{(z+t+x)(t+x+y)(x+y+z)}} \leq \sum \frac{(\frac{y+z}{x+y+z}+\frac{z+t}{z+t+x}+\frac{t+y}{x+y+t}) }{6}=\frac{4}{3}$

The positive real numbers $a,b, c,d$ satisfy the equality $ \frac {1}{a^3+1} + \frac {1}{b^3+1} + \frac {1}{c^3+1} + \frac{ 1}{d^3+1} = 3 $ . Prove the inequality $$abc+bcd+cda+dab\le \frac {4}{3} $$Old? Where?

sqing wrote: The positive real numbers $a,b, c,d$ satisfy the equality $ \frac {1}{a^3+1} + \frac {1}{b^3+1} + \frac {1}{c^3+1} + \frac{ 1}{d^3+1} = 3 $ . Prove the inequality $$abc+bcd+cda+dab\le \frac {4}{3} $$Old? Where? The problem is the same, but another substitution, $a=\sqrt[3]{\frac{x}{y+z+t}}$

that's what i call magic sbstitutions ! really marvellous !

Muradjl wrote: that's what i call magic sbstitutions ! really marvellous ! There is no any magic, it follows directly from $\frac{1}{1+a}=\frac{y+z+t}{x+y+z+t}$, $y+z+t$ because $\frac{1}{1+a}<1$

Another way to get the same substitution is to consider it's "dual". \[\frac {1}{a+1} + \frac {1}{b+1} + \frac {1}{c+1} + \frac{ 1}{d+1} = 3\iff\frac {a}{a+1} + \frac {b}{b+1} + \frac {c}{c+1} + \frac{d}{d+1} = 1\]Setting $w = \frac{a}{1+a}$, we get $w+x+y+z=1$ and $a = \frac{w}{1-w} = \frac{w}{x+y+z}$.

Generalization: Let $n\in\mathbb{N}_{\geq 3}$, $a_i \in \mathbb{R^+}\ \forall 1\leq i\leq n$, such that $\textstyle{\sum_{i=1}^n (1+a_i)^{-1} = n-1}$. Then prove that: \[\sum_{1\leq i < j < k\leq n} \sqrt[3]{a_ia_ja_k} \leq \frac{n(n-2)}{6}.\] This is the case where $n=4$, RMO Mumbai 2016/P2 is the case where $n=3$.

Proof

Vrangr wrote: Generalization: Let $n\in\mathbb{N}_{\geq 3}$, $a_i \in \mathbb{R^+}\ \forall 1\leq i\leq n$, such that $\textstyle{\sum_{i=1}^n (1+a_i)^{-1} = n-1}$. Then prove that: \[\sum_{1\leq i < j < k\leq n} \sqrt[3]{a_ia_ja_k} \leq \frac{n(n-2)}{6}.\] Let $b_i=\frac 1{1+a_i}.$ Then $a_i=\frac{1-b_i}{b_i}$ \begin{align*}\sum_{1\leq i < j < k\leq n} \sqrt[3]{a_ia_ja_k} &=\frac 12\sum_{1\leq i < j < k\leq n} \left(\sqrt[3]{\frac{1-b_i}{b_k}\cdot\frac{1-b_j}{b_i}\cdot\frac{1-b_k}{b_j}}+\sqrt[3]{\frac{1-b_i}{b_j}\cdot\frac{1-b_j}{b_k}\cdot\frac{1-b_k}{b_i}} \right)\\&\le\frac 16\sum_{1\leq i < j < k\leq n}\left({\frac{1-b_i}{b_k}+\frac{1-b_j}{b_i}+\frac{1-b_k}{b_j}}+{\frac{1-b_i}{b_j}+\frac{1-b_j}{b_k}+\frac{1-b_k}{b_i}} \right)\\&=\frac{n-2}6\sum_{i=1}^n\sum_{j\neq i}\frac{1-b_j}{b_i}= \frac{n(n-2)}{6}.\Box\end{align*}