<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] other solution:
since $xyz=1$,we only prove
$$\dfrac{1}{1+x+y}+\dfrac{1}{1+y+z}+\dfrac{1}{1+x+z}\le 1$$
let
$$x=\dfrac{a^2}{bc},y=\dfrac{b^2}{ac},z=\dfrac{c^2}{ab}$$
so
$$\Longrightarrow  \dfrac{1}{1+y+z}=\dfrac{1}{1+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}}=\dfrac{abc}{abc+b^3+c^3}$$
so this inequality we only prove 
$$\Longleftrightarrow \dfrac{abc}{abc+a^3+c^3}+\dfrac{abc}{abc+b^3+c^3}+\dfrac{abc}{abc+a^3+b^3}\le 1$$
note
$$a^3+b^3\ge ab(a+b)\Longrightarrow a^3+b^3+abc\ge ab(a+b+c)$$
$$\Longrightarrow \dfrac{abc}{a^3+b^3+abc}\le\dfrac{abc}{ab(a+b+c)}=\dfrac{c}{a+b+c}$$
so
$$\sum_{cyc}\dfrac{abc}{a^3+b^3+abc}\le\sum_{cyc}\dfrac{c}{a+b+c}=1$$
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