# A Cycling Inequality with Integrals

### Proof

Note that

$\displaystyle\Omega'(t)=\int_{0}^{1}\frac{(1-x^2)(-1)x^2}{(1+tx^2+x^4)^2}dx\lt 0,$

for $t\ge 2,\;$ making $\Omega(t)\;$ strictly decreasing on $[2,\infty).\;$ Further

\displaystyle\begin{align} \Omega(2)&=\int_{0}^{1}\frac{1-x^2}{1+2x^2+x^4}dx\\ &=\int_{0}^{1}\left[\frac{2}{(1+x^2)^2}-\frac{1}{1+x^2}\right]dx. \end{align}

But

\begin{align} \frac{\pi}{4} &= \int_{0}^{1}\frac{dx}{1+x^2}\\ &=\frac{x}{1+x^2}\bigg|^{1}_{0}+\int_{0}^{1}\frac{x(2x)}{(1+x^2)^2}dx\\ &=\frac{1}{2}+\int_{0}^{1}\frac{2(x^2+1)-2}{(1+x^2)^2}dx\\ &=\frac{1}{2}+\frac{\pi}{4}-\Omega(2), \end{align}

implying $\displaystyle\Omega(2)=\frac{1}{2}\;$ such that, for $t\gt 2,\;$ $0\lt\Omega(t)\lt\displaystyle\frac{1}{2}.\;$ It follows that

\displaystyle\begin{align} 2bc\Omega(a)+2ca\Omega(b)+2ab\Omega(c)&\lt bc+ca+ab\\ &\le\frac{b^2+c^2}{2}+\frac{c^2+a^2}{2}+\frac{a^2+b^2}{2}\\ &=a^2+b^2+c^2. \end{align}

### Acknowledgment

Dan Sitaru has kindly posted the above problem from the Romanian Mathematical Magazine (and his book Math Accent), with a proof by Ravi Prakash (India), at the CutTheKnotMath facebook page.

Note that the penultimate step in the proof could be shortened by noticing that

$\displaystyle\int\frac{1-x^2}{(1+x^2)^2}dx=\frac{x}{1+x^2}+C.$