# An Inequality with Absolute Values

### Proof 1

Since $a\in (-1,1),\;$ $a^2\lt 1,\;$ $1-a^2\gt 0,\;$ $\displaystyle \frac{1}{1-a^2}\gt 0.\;$ Similarly, $\displaystyle \frac{1}{1-b^2}\gt 0.\;$

By the AM-GM inequality,

(1)

$\displaystyle\frac{1}{1-a^2}+\frac{1}{1-b^2}\ge 2\sqrt{\frac{1}{1-a^2}\cdot\frac{1}{1-b^2}}.$

However,

\displaystyle\begin{align} (1-ab)^2&=1-2ab+a^2b^2\\ &\ge 1-(a^2+b^2)-a^2b^2\\ &=(1-a^2)(1-b^2), \end{align}

so that $\displaystyle\frac{1}{(1-a^2)(1-b^2)}\ge\frac{1}{(1-ab)^2}.\;$ This, together with (1), yields

$\displaystyle \frac{1}{1-a^2}+\frac{1}{1-b^2}\ge 2\sqrt{\frac{1}{(1-ab)^2}}=\frac{2}{1-ab}.$

So too,

$\displaystyle \frac{|c|}{1-a^2}+\frac{|c|}{1-b^2}\ge \frac{2|c|}{1-ab}.$

Similarly,

$\displaystyle \frac{|a|}{1-b^2}+\frac{|a|}{1-c^2}\ge \frac{2|a|}{1-bc}\\ \displaystyle \frac{|b|}{1-c^2}+\frac{|b|}{1-a^2}\ge \frac{2|b|}{1-ca}.$

Adding the three gives the required inequality. The equality is achieved for $a=b=c.$

### Proof 2

Using Bergström's inequality and, subsequently, the obvious $b^2+c^2\ge 2bc,$

\displaystyle\begin{align} \frac{1}{1-b^2}+\frac{1}{1-c^2}&\ge\frac{(1+1)^2}{2-b^2-c^2}\\ &\ge\frac{4}{2-2bc}\\ &=\frac{2}{1-bc}, \end{align}

so that

$\displaystyle \frac{|a|}{1-b^2}+\frac{|a|}{1-c^2}\ge\frac{2|a|}{1-bc}.$

Similarly,

$\displaystyle \frac{|b|}{1-c^2}+\frac{|b|}{1-a^2}\ge \frac{2|b|}{1-ca},\\ \displaystyle \frac{|c|}{1-a^2}+\frac{|c|}{1-b^2}\ge \frac{2|c|}{1-ab}.$

Adding the three gives the required inequality. The equality is achieved for $a=b=c.$

### Acknowledgment

Dan Sitaru has kindly posted the above problem (from his book Math Accent), with a solution (Proof 1), at the CutTheKnotMath facebook page. He later added another solution (Proof 2) by Kevin Soto Palacios.