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If $a,b,c>0$, are positive real numbers such that $a^2+b^2+c^2=1$ then, the following Inequalities hold:

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

$\displaystyle 1 \le \frac{a}{1+ab}+\frac{b}{1+bc}+\frac{c}{1+ac} \le \frac{3\sqrt{3}}{4}$

Homogenizing the first inequality as, $\displaystyle \frac{a}{a^2+b^2+c^2-ab}+\frac{b}{a^2+b^2+c^2-bc}+\frac{c}{a^2+b^2+c^2-ac} \le \frac{3\sqrt{3}}{2\sqrt{a^2+b^2+c^2}}$

and noting the cyclic symmtery I tried Rearrangement Inequalities Inequalities.

Using $a^2+b^2+c^2-ab = \frac{1}{4}(a+b)^2+\frac{3}{4}(a-b)^2+c^2 \ge \frac{1}{4}(a+b)^2+c^2 \ge (a+b)c$

$\displaystyle \sum_{cyc} \frac{a}{a^2+b^2+c^2-ab} \le \sum_{cyc} \frac{a}{ac+bc}$ might help, but I couldn't get anywhere with it.

Thank you!

Michael Rozenberg
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r9m
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3 Answers3

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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 $$3p^7-15p^6+27p^5-247p^4 +717p^3-1953p^2+621p-81\le 0(p\ge \sqrt{3})$$

math110
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The left inequalities we can prove by Holder:

  1. $$\left(\sum_{cyc}\frac{a}{1-ab}\right)^2\sum_{cyc}a(1-ab)^2\geq(a+b+c)^3.$$

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;

  1. $$\left(\sum_{cyc}\frac{a}{1+ab}\right)^2\sum_{cyc}a(1+ab)^2\geq(a+b+c)^3.$$

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.

Michael Rozenberg
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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 $$1\leq \sum_{cyc} \frac{a}{1-ab}$$

Fabien
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  • Is that Holder's Inequality with negative index ? Could you link me to a proof of this negative index inequality ? .. Thank you :) – r9m Jun 02 '14 at 02:10
  • When I googled it, I find : downloads.hindawi.com/journals/ijmms/2001/324106.pdf – Fabien Jun 02 '14 at 02:22
  • the link says " .. the situation where all but one of the $p_{\alpha}$’s are real numbers lying in the interval $(0,1)$ hence the one is forced to be negative in order that $\sum \frac{1}{p_{\alpha}} = 1$" is incorrect :O .. but then again that is not our case .. we have one index $\frac{1}{1/2} > 1$ outside the interval $(0,1)$ unfortunately the link doesn't say anything about that case (please tell me if I missed anything crucial) .. :) – r9m Jun 02 '14 at 03:19
  • Yeah, the pdf doesn't give a clear statement, but I'm afraid it's wrong. I'll edit, thanks for the comment ;) – Fabien Jun 02 '14 at 10:39