# A Cyclic Inequality in Three Variables XXIV

### Lemma

For $x\gt 0,$

$\displaystyle \frac{1+x^2}{1+x}\ge 2(\sqrt{2}-1).$

Proof by algebra

\begin{align} &\left(x-(\sqrt{2}-1)\right)^2\ge 0&\,\Longleftrightarrow\\ &x^2-2(\sqrt{2}-1)(\sqrt{2}-1)^2\ge 0&\,\Longleftrightarrow\\ &x^2-2(\sqrt{2}-1)(3-2\sqrt{2})\ge 0&\,\Longleftrightarrow\\ &1+x^2\ge2\sqrt{2}+2x\sqrt{2}-2-2x&\,\Longleftrightarrow\\ &1+x^2\ge (1+x)(2\sqrt{2}-2)&\,\Longleftrightarrow\\ &\frac{1+x^2}{1+x}\ge 2(\sqrt{2}-1). \end{align}

Proof by calculus

If $\displaystyle f(x)=\frac{1+x^2}{1+x},\,$ $\displaystyle f'(x)=\frac{x^2+2x-1}{(1+x)^2},\,$ with the only positive zero at $x=\sqrt{2}-1,\,$ where $f(\sqrt{2}-1)=2(\sqrt{2}-1).\,$ Further, $\displaystyle f''(x)=\frac{4}{(1+x)^3}\ge 0,\,$ implying $f(x)\ge 2(\sqrt{2}-1).$

### Solution

By Lemma,

$\displaystyle \frac{(1+a^2)b^2}{1+a}\ge 2b^2(\sqrt{2}-1)\,\text{and}\\ \displaystyle \frac{(1+b^2)a^2}{1+b}\ge 2a^2(\sqrt{2}-1),$

so that, given that the sequence $(2,2,0)\,$ majorizes $(2,1,1),$

\displaystyle\begin{align} \frac{a^2b^2(1+a^2)(1+b^2)}{(1+a)(1+b)}&\geq 4(\sqrt{2}-1)^2a^2b^2\;\text{and, subsequently,}\\ \sum_{cycl} \frac{a^2b^2(1+a^2)(1+b^2)}{(1+a)(1+b)}&\geq 4(\sqrt{2}-1)^2\sum_{cycl} a^2 b^2\\ &\overbrace{\geq}^{Muirhead}4(2+1-2\sqrt{2})\sum_{cycl} a^2bc\\ &= 4(3-2\sqrt{2})abc(a+b+c). \end{align}

Inequality is attained for $a=b=c=\sqrt{2}-1.$

### Acknowledgment

Dan Sitaru has kindly posted the problem from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page and later communicated the above proof in a LaTeX file.