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Let $a,b,c \ge0$, prove the on equality: $${\frac {1+{a}^{3}}{1+a{b}^{2}}}+{\frac {1+{b}^{3}}{1+b{c}^{2}}}+{ \frac {1+{c}^{3}}{1+c{a}^{2}}}\ge 3 $$

I tried: $$LHS = \sum\frac 1{1+ab^2}+\sum \frac {a^4}{a+a^2b^2} \ge\frac 9{3+\sum ab^2} + \frac {(a^2+b^2+c^2)^2}{\sum a+ \sum (ab)^2}\ge...$$ but it seem like useless.

Look like better inequality is $ \left( 1+{a}^{3} \right) \left( 1+{b}^{3} \right) \left( 1+{c}^{3} \right) \ge \left( 1+a{b}^{2} \right) \left( 1+b{c}^{2} \right) \left( 1+c{a}^{2} \right)$ (but still can't prove it)

Xeing
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    Have you heard of Rearrangement Inequality? When you have $a \ge b \ge c$, then you have $a^2+b^2+c^2 \ge ab+bc+ca$, or $a^3 \ge a^2bc \ge b^2a \ge b^2c $ – Inceptio May 26 '13 at 14:54
  • Yes, but not familiar with it. You can solve the problem by Rearrangement Inequality, I think I 'll understand the solution. – Xeing May 26 '13 at 15:08
  • The last inequality $\left( 1+{a}^{3} \right) \left( 1+{b}^{3} \right) \left( 1+{c}^{3} \right) \ge \left( 1+a{b}^{2} \right) \left( 1+b{c}^{2} \right) \left( 1+c{a}^{2} \right)$ is direct from Re-arrangement inequality. I'm still working on the first one. – Inceptio May 26 '13 at 15:09
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    So, the problem is so easy: ${\frac {1+{a}^{3}}{1+a{b}^{2}}}+{\frac {1+{b}^{3}}{1+b{c}^{2}}}+{ \frac {1+{c}^{3}}{1+c{a}^{2}}}\ge 3 \sqrt[3]{{\frac {1+{a}^{3}}{1+a{b}^{2}}}\cdot{\frac {1+{b}^{3}}{1+b{c}^{2}}}\cdot{ \frac {1+{c}^{3}}{1+c{a}^{2}}}} \ge 3$ – Xeing May 26 '13 at 15:19
  • But can you make more detail how to use Rearrangement Inequality – Xeing May 26 '13 at 15:22
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    It is quite direct. You just need to notice that. When you expand the inequalities, you get :

    $a^3b^3c^3+a^3b^3+a^3c^3+a^3+b^3c^3+b^3+c^3+1 \ge a^3b^3c^3+a^3b^2c+a^2bc^3+a^c+ab^3c^2+ab^2+bc^2+1$, which is direct from Re-arrangement inequality.

     – Inceptio May 26 '13 at 15:33

2 Answers2

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Without loss of generality, let $a\geqslant b\geqslant c\geqslant0$. The target inequality is equivalent to: $$\left(\frac{1+{a}^{3}}{1+a{b}^{2}}-1\right)+\left(\frac{1+{b}^{3}}{1+b{c}^{2}}-1\right)+\left(\frac{1+{c}^{3}}{1+c{a}^{2}}-1\right)\geqslant0$$ That is \begin{align*} a\frac{a^2-b^2}{1+ab^2}+b\frac{b^2-c^2}{1+bc^2}+c\frac{c^2-a^2}{1+ca^2}\geqslant0\Leftrightarrow\\ a\frac{a^2-b^2}{1+ab^2}+b\frac{b^2-c^2}{1+bc^2}+c\left(\frac{c^2-b^2}{1+ca^2}+\frac{b^2-a^2}{1+ca^2}\right)\geqslant0\Leftrightarrow\\ (a^2-b^2)\left(\frac{a}{1+ab^2}-\frac{c}{1+ca^2}\right)+(b^2-c^2)\left(\frac{b}{1+bc^2}-\frac{c}{1+ca^2}\right)\geqslant0 \end{align*} Think of the numerators of $\frac{a}{1+ab^2}-\frac{c}{1+ca^2}$ and $\frac{b}{1+bc^2}-\frac{c}{1+ca^2}$ are $$(a-c)+ac(a^2-b^2)\geqslant0~\mbox{and}~(b-c)+bc(a^2-c^2)\geqslant0$$Hence we complete the proof.

Easy
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By AM-GM and Holder we obtain: $$\sum_{cyc}\frac{1+a^3}{1+ab^2}\geq3\sqrt[3]{\frac{\prod\limits_{cyc}(1+a^3)}{\prod\limits_{cyc}(1+ab^2)}}=3\sqrt[3]{\frac{\sqrt[3]{\prod\limits_{cyc}(1+a^3)^3}}{\prod\limits_{cyc}(1+ab^2)}}=$$ $$=3\sqrt[3]{\frac{\sqrt[3]{\prod\limits_{cyc}(1+a^3)(1+b^3)^2}}{\prod\limits_{cyc}(1+ab^2)}}\geq3\sqrt[3]{\frac{\prod\limits_{cyc}(1+ab^2)}{\prod\limits_{cyc}(1+ab^2)}}=3.$$ Done!

Michael Rozenberg
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