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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.

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  • $\begingroup$ $$\frac{ab}{4a + b + 4c} + \frac{cd}{4c + d + 4a} = \frac{(a + c)(4ab + bd + 4cd)}{(4a + b + 4c)(4c + d + 4a)}$$ $\endgroup$ – Lê Thành Đạt Jun 28 at 14:57
  • $\begingroup$ I thought the Chebyshev inequality would work because of the above equation but it wasn't. $\endgroup$ – Lê Thành Đạt Jun 28 at 15:25
  • $\begingroup$ I think, Chebyshov does not help here, but you are welcome to show your proof. $\endgroup$ – Michael Rozenberg Jun 28 at 15:26
  • $\begingroup$ No, it was only an idea, not a full solution. Can I ask about the source of the problem? I'm really interested in this equality in particular. $\endgroup$ – Lê Thành Đạt Jun 28 at 15:28
  • $\begingroup$ I don't remember exactly. Maybe from AoPS. $\endgroup$ – Michael Rozenberg Jun 28 at 15:29

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