For $a,b,c,d>0$ such that $abcd\ge 1$. Prove that $$\frac{1}{3a+2b+c+6}+\frac{1}{3b+2c+d+6}+\frac{1}{3c+2d+a+6}+\frac{1}{3d+2a+b+6} \leq \frac{1}{3}$$
My attempts: By AM-GM $$3a+2b+c+6\geq12\sqrt[12]{a^3b^2c}$$ and I have to prove that $$\frac{1}{\sqrt[12]{a^3b^2c}}+\frac{1}{\sqrt[12]{b^3c^2d}}+\frac{1}{\sqrt[12]{c^3d^2a}}+\frac{1}{\sqrt[12]{d^3a^2b}}\leq4,$$
But it's wrong.
Let $a=kx$, $b=ky$, $c=kz$ and $d=kt$, where $k>0$ and $xyzt=1$.
Thus, $$k^4xyzt\geq1,$$ which gives $k\geq1$ and we obtain: $$\sum_{cyc}\frac{1}{3a+2b+c+6}=\frac{1}{k(3x+2y+z)+6}\leq\sum_{cyc}\frac{1}{3x+2y+z+6}.$$ Id est, it's enough to prove that: $$\sum_{cyc}\frac{1}{3x+2y+z+6}\leq\frac{1}{3}$$ or $$\sum_{cyc}\left(\frac{1}{3x+2y+z+6}-\frac{1}{6}\right)\leq\frac{1}{3}-\frac{2}{3}$$ or $$\sum_{cyc}\frac{3x+2y+z}{3x+2y+z+6}\geq2.$$ Now, by C-S we obtain: $$\sum_{cyc}\frac{3x+2y+z}{3x+2y+z+6}=\sum_{cyc}\frac{(3x+2y+z)^2}{(3x+2y+z)^2+6(3x+2y+z)}\geq$$ $$\geq\frac{\left(\sum\limits_{cyc}(3x+2y+z)\right)^2}{\sum\limits_{cyc}((3x+2y+z)^2+6(3x+2y+z))}=$$ $$=\frac{36(x+y+z+t)^2}{\sum\limits_{cyc}(14x^2+16xy+6xz)+36(x+y+z+t)}$$ and it's enough to prove that $$9(x+y+z+t)^2\geq\sum_{cyc}(7x^2+8xy+3xz+18x)$$ or $$9(x+y+z+t)^2+2(xz+yz)\geq\sum_{cyc}(7x^2+8xy+4xz+18x).$$ Now, let $x+y+z+t=4u$, $xy+xz+yz+xt+yz+zt=6v^2$, $xyz+xyt+xzt+yzt=4w^3$ and $xyzt=t^4$, where $t>0$.
Thus, since by AM-GM $$xt+yz\geq2\sqrt{xyzt}=2t^2,$$ it's enough to prove that $$9\cdot16u^2+4t^2\geq7(16u^2-12v^2)+8\cdot6v^2+18\cdot4ut$$ or $$8u^2+9v^2+t^2\geq18ut,$$ which is true by AM-GM: $$8u^2+9v^2+t^2\geq18\sqrt[18]{u^{16}v^{18}t^2}=18\sqrt[18]{u^{16}v^{8}\cdot v^{10}t^2}\geq$$ $$\geq18\sqrt[18]{u^{16}\cdot u^2w^6\cdot t^{12}}\geq18\sqrt[18]{u^{18}t^{18}}=18ut$$ and we are done!