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.

You could approach it as follows:
$$
\sum_{cyc} \frac{abc}{1+bc}=\sum_{cyc} a\left(1-\frac{1}{1+bc}\right)=\sum_{cyc} \left(a-\frac{a}{1+bc}\right)=1-\sum_{cyc} \frac{a}{1+bc}
$$
So the inequality is equivalent to:
$$
1-\sum_{cyc} \frac{a}{1+bc}\le\frac{1}{17}\iff\frac{16}{17}\le\sum_{cyc} \frac{a}{1+bc}
$$
Using CS, this can be reduced to prove the following:
$$
\left(\sum_{cyc} \frac{a}{1+bc}\right)\cdot\left(\sum_{cyc} a(1+bc)\right)\ge(a+b+c+d)^2=1\iff\sum_{cyc} \frac{a}{1+bc}\ge\frac{1}{\sum_{cyc} a(1+bc)}=\frac{1}{a+b+c+d+abc+bcd+cda+dab}=\frac{1}{1+abc+bcd+cda+dab}
$$
So if
$$
\frac{1}{1+abc+bcd+cda+dab}\ge\frac{16}{17}\iff abc+bcd+cda+dab\le\frac{1}{16}
$$
is true, the original inequality would be true as well.
Edit:
The inequality
$$
abc+bcd+cda+dab\le\frac{1}{16}
$$
is true due to Maclaurin's inequality , which, in a special case, states that:
$$
\left(\frac{abc+bcd+cda+dab}{4}\right)^{\frac13}\le\frac{a+b+c+d}{4}=\frac{1}{4}\iff abc+bcd+cda+dab\le\frac{1}{16}
$$
And your inequality is proven.
