An Inequality Problem $1 \le \frac{a}{1-ab}+\frac{b}{1-bc}+\frac{c}{1-ac} \le \frac{3\sqrt{3}}{2}$

we can prove $$\dfrac{x}{1+xy}\le\dfrac{3\sqrt{3}x(x+y+2z)}{4(x^2+y^2+z^2+3(xy+yz+xz))}$$ so $$\sum_{cyc}\dfrac{x}{1+xy}\le \sum_{cyc}\left(\dfrac{3\sqrt{3}x(x+y+2z))}{4(x^2+y^2+z^2+3(xy+yz+xz)}\right)=\dfrac{3\sqrt{3}}{4}$$

I have an other ugly methods:

this problem is equivalent:$a^2+b^2+c^3=3$,then we have $$\dfrac{a}{ab+3}+\dfrac{b}{bc+3}+\dfrac{c}{ca+3}\le\dfrac{3}{4}$$ $$\Longleftrightarrow 4abc\sum_{cyc}ab+12\sum_{cyc}ab^2+36abc+36\sum_{cyc}a \le 3a^2b^2c^2+9abc\sum_{cyc}a+27\sum_{cyc}ab+81$$ use this follow well know reslut $$ab^2+bc^2+ca^2\le\dfrac{4}{27}(a+b+c)^3-abc$$ so we only $$4abc\sum_{cyc}ab+12\left(\dfrac{4}{27}(a+b+c)^3-abc\right) +36abc+36\sum_{cyc}a\le 3a^2b^2c^2+9abc\sum_{cyc}a+27\sum_{cyc}ab+81$$ let $$p=a+b+c,q=ab+bc+ac,r=abc,p^2-2q=3$$ use AM-GM inequality we have $$\dfrac{(p^2-3)^2}{4}=q^2\ge 3qr$$ so we consider $$f(r)=3r^2-(2p^2-9p+18)r-\dfrac{16}{9}p^3+\dfrac{27}{2}p^2-36p +\dfrac{81}{2}\ge 0$$ so $$f'(r)=6r-2p^2+9p-18\le\dfrac{(p^2-3)^2}{2p}-2p^2+9p-18= \dfrac{(p-1)(p-3)(p^2+2)-18}{2p}\le 0$$ so $$f(r)\ge f\left(\dfrac{(p^2-3)^2}{12}\right)=\dfrac{(p-3)(3p^7-15p^6+27p^5-247p^4 +717p^3-1953p^2+621p-81)}{144p^2}$$ and we can prove $$3p^7-15p^6+27p^5-247p^4 +717p^3-1953p^2+621p-81\le 0(p\ge \sqrt{3})$$