<s>[INST] Show that: $\frac{1}{1+\log_ab+\log_bc}+\frac{1}{1+\log_bc+\log_ca}+\frac{1}{1+\log_ca+\log_ab}\leq1$ 
Let $a, b, c>1$. Show that:
  $$\frac{1}{1+\log_ab+log_bc}+\frac{1}{1+\log_bc+\log_ca}+\frac{1}{1+\log_ca+\log_ab}\leq1.$$

My attempt: 
We noted $\log_bc=x, \log_ca=y, \log_ab=z$ with $xyz=1$ and we have reduced inequality at this time:$$2(x+y+z)\leq xy^2+x^2y+yz^2+y^2z+xz^2+x^2z.$$
Does anyone have idea how it goes? Thank You!
I found a simple continuation:
$$2(x+y+z)\leq xy^2+x^2y+yz^2+y^2z+xz^2+x^2z\iff$$
$$\iff2(x+y+z)\leq xy(x+y)+yz(y+z)+zx(z+x)\iff$$
$$\iff2(x+y+z)\leq (xy+yz+zx)(x+y+z)-3xyz\iff$$
$$\iff3\leq(x+y+z)(xy+yz+zx-2).$$
The last inequality follows easily using:
$$3\leq x+y+z$$
$$1=3(xyz)^{\frac{2}{3}} -2\leq xy+yz+zx-2.$$
These last inequality is obtained by applying AM-GM inequality.
 [/INST] Following OP's simplification of the inequality:
$\displaystyle \begin{align} xy^2+x^2y+yz^2+y^2z+xz^2+x^2z &= \sum\limits_{cyc} (x^2y + xy^2) \\ &=\sum\limits_{cyc} (x^{6/3}y^{3/3} + x^{3/3}y^{6/3}) \\ & \ge \sum\limits_{cyc} (x^{5/3}y^{4/3} + x^{4/3}y^{5/3}) \tag{1}\\ &=\sum\limits_{cyc} x^{5/3}(y^{4/3} + z^{4/3})\tag{2} \\ & \ge 2\sum\limits_{cyc} x^{5/3}y^{2/3}z^{2/3} \tag{3}\\ &= 2\sum\limits_{cyc} x \end{align}$
Justification of steps:
$(1) \, (x^{6/3}y^{3/3} + x^{3/3}y^{6/3}) - (x^{5/3}y^{4/3} + x^{4/3}y^{5/3}) = xy(x^{1/3}+y^{1/3})(x^{1/3}-y^{1/3})^2 \ge 0$
$(2)$ Rearranging the cyclic summation.
$(3) \, y^{4/3} + z^{4/3} \ge 2y^{2/3}z^{2/3}$ and put $xyz = 1$.
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