How to prove$\sum\frac{a}{a^2+a+1}\leq\sum\frac{1}{a^2+a+1}$ If I asked the obvious because of my lack of knowledge, then I am sorry!
If $abcd=1,a,b,c,d>0$;prove:$$\sum_\limits{cyc}\frac{a}{a^2+a+1}\leq\sum_\limits{cyc}\frac{1}{a^2+a+1}$$
I use $f(x)=\frac{x-1}{x^2+x+1}$,it without success;I want to use$a=\frac{yz}{x^2},b=\frac{zt}{y^2},c=\frac{tx}{z^2},d=\frac{xy}{t^2},$and the Order inequality,but I'm not sure if it's right.Could anyone help me? Thanks a lot!

$\sum\limits_{cyc}\frac{1-a}{1+a+a^2}=\sum\limits_{cyc}\left(\frac{1-a}{1+a+a^2}+\frac{1}{3}\ln{a}\right)$.
Let $f(x)=\frac{1-x}{x^2+x+1}+\frac{1}{3}\ln{x}$.
Hence, $f'(x)=\frac{(x-1)(x^3+6x^2+3x-1)}{3x(x^2+x+1)^2}$.
Which says that there is unique $0<x_1<1$, for which $f(x)\geq0$ for all $x\geq x_1$. 
Easy to see that $x_1=0.0779...$, 
which says that for $\min\{a,b,c,d\}\geq0.08$ our inequality is true.
Let $a<0.08$ and $g(x)=\frac{1-x}{1+x+x^2}$.
Hence, $g'(x)=\frac{x^2-2x-2}{(1+x+x^2)^2}$, which says that $\min\limits_{x>0}g=g(1+\sqrt3)$ 
and $g$ is a decreasing function on $(0,0.08]$.
Id est, $\sum\limits_{cyc}\frac{1-a}{1+a+a^2}\geq g(0.08)+3g(1+\sqrt3)=0.38...>0$
Done!
