Let $a$, $b$, $c$ and $d$ be non-negative numbers such that $\sum\limits_{sym}ab\neq0$. Prove that: $$\frac{ab}{4a+b+4c}+\frac{bc}{4b+c+4d}+\frac{cd}{4c+d+4a}+\frac{da}{4d+a+4b}\leq\frac{a+b+c+d}{9}$$
The equality occurs also for $a=b=0$ and $d=2c$.
There is a similar inequality for three variables:
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that: $$\frac{ab}{4a+b+4c}+\frac{bc}{4b+c+4a}+\frac{ca}{4a+c+4b}\leq\frac{a+b+c}{9},$$ which we can prove by the following reasoning.
By C-S $$\sum_{cyc}\frac{ab}{4a+b+4c}=\sum_{cyc}\frac{ab}{b+2c+2(c+2a)}\leq\sum_{cyc}\frac{ab}{9}\left(\frac{1^2}{b+2c}+\frac{2^2}{2(c+2a)}\right)=$$ $$=\frac{1}{9}\sum_{cyc}\left(\frac{ab}{b+2c}+\frac{2ab}{c+2a}\right)=\frac{1}{9}\sum_{cyc}\left(\frac{bc}{c+2a}+\frac{2ab}{c+2a}\right)=\frac{a+b+c}{9},$$ but it does not help for a proof of the starting inequality (at least I don't see, how it helps).
I tried also BW, but we get there something, which impossible to kill during a competition.