<s>[INST] Let $a,b,c,d>0$ and $a+b+c+d=1$. Prove that $\frac{abc}{1+bc}+\frac{bcd}{1+cd}+\frac{cda}{1+ad}+\frac{dab}{1+ab}\le \frac{1}{17}$ Let $a,b,c,d>0$ and $a+b+c+d=1$. Prove that $\dfrac{abc}{1+bc}+\dfrac{bcd}{1+cd}+\dfrac{cda}{1+ad}+\dfrac{dab}{1+ab}\le \dfrac{1}{17}$  
My attempt:  
I figured out that if each of the element could be like $\dfrac{abc}{1+bc}\le \dfrac{1}{68}$ then we would be done.  
From a little manipulation we get, $\dfrac{1}{bc}+1\le 68a$ , $\dfrac{1}{cd}+1\le 68b$ , $\dfrac{1}{da}+1\le 68c$ , $\dfrac{1}{ab}+1\le 68d$. Summing then we get, 
$\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{cd}+\dfrac{1}{da}\le 64$  
But, I suppose my first assumption could be wrong, if not then please help me further and if so then please help with the solution. Thank you.
 [/INST] Here is my approach. I am stuck at a place but I hope this is a good method and I don't reach a dead end. The LHS is equivalent to $\frac{a}{1/bc+1}+\frac{b}{1/cd+1}+\frac{c}{1/da+1}+\frac{d}{1/ab+1}$. From AM-GM, we have $1+1/bc>2/\sqrt{bc}$. Inverse the inequality with inverting the inequality sign and then multiply by $a$ and write the other 3 terms. Then add all the terms. Denote our original expression as P. We have $P<a\sqrt{bc}+b\sqrt{cd}+c\sqrt{da}+d\sqrt{ab}/2$. Which itself is less than $ab+bd+ab+ac+bc+bd+cd+ac/4$. Now I am working to prove this less than 1/17, hope this helps.
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