An Inequality Problem $1 \le \frac{a}{1-ab}+\frac{b}{1-bc}+\frac{c}{1-ac} \le \frac{3\sqrt{3}}{2}$
Solution 1:
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 $$3p^7-15p^6+27p^5-247p^4 +717p^3-1953p^2+621p-81\le 0(p\ge \sqrt{3})$$
Solution 2:
Please consider the following as a comment, I'm not able to say if the Holder equality is valid in that case.
RHS for the first inequality : $$1-ab=1+\frac{(a-b)^2-a^2-b^2}{2}=\frac12+\frac{(a-b)^2+c^2}{2}\geq \frac{1+c^2}{2}$$ so $$\sum_{cyc} \frac{a}{1-ab}\leq \sum_{cyc} \frac{a}{\frac{1+c^2}{2}}$$ We then use the Hölder's inequality : $$\sum_{cyc} \frac{a}{1-ab}\leq \left(\sum_{cyc} a^2\right)^{\frac12}\left(\sum_{cyc} \frac{1+c^2}{2}\right)^{-1}\left(\sum_{cyc} 1\right)^{\frac32}=\frac{3\sqrt{3}}{2}$$
LHS for the first inequality : $$1-ab\leq 1$$ so $$\sum_{cyc} \frac{a}{1-ab}\geq \sum_{cyc} a$$ Or : $$1=\sum_{cyc} a^2\leq \left(\sum_{cyc} a\right)^{2}$$ so $$1\leq \sum_{cyc} \frac{a}{1-ab}$$
Solution 3:
The left inequalities we can prove by Holder:
- $$\left(\sum_{cyc}\frac{a}{1-ab}\right)^2\sum_{cyc}a(1-ab)^2\geq(a+b+c)^3.$$
Thus, it remains to prove that $$(a+b+c)^3\geq \sum_{cyc}a(1-ab)^2$$ or $$(a+b+c)^3\geq(a+b+c)(a^2+b^2+c^2)-2\sum_{cyc}a^2b+\sum_{cyc}a^3b^2$$ or $$\sum_{cyc}(4a^2b+2a^2c+2abc-a^3b^2)\geq0,$$ for which it's enough to prove that $$\sum_{cyc}a^2c\geq\sum_{cyc}a^3b^2$$ or $$\sum_{cyc}ab^2\sum_{cyc}a^2\geq\sum_{cyc}a^3b^2,$$ which is obvious;
- $$\left(\sum_{cyc}\frac{a}{1+ab}\right)^2\sum_{cyc}a(1+ab)^2\geq(a+b+c)^3.$$
Thus, it remains to prove that $$(a+b+c)^3\geq \sum_{cyc}a(1+ab)^2$$ or $$(a+b+c)^3\geq(a+b+c)(a^2+b^2+c^2)+2\sum_{cyc}a^2b+\sum_{cyc}a^3b^2$$ or $$\sum_{cyc}(2a^2c+2abc-a^3b^2)\geq0,$$ for which it's enough to prove that $$\sum_{cyc}a^2c\geq\sum_{cyc}a^3b^2,$$ which is obvious.
The right inequalities are true, but I still have no a nice proof.