# A Cycling Inequality with Integrals II

### Proof

Introduce $u=x^a,\;$ $v=x^b,\;$ $w=x^c.\;$ For $x\in [0,1],\;$ $u,v,w\in [0,1],\;$ such that also $1-u,1-v,1-w\in [0,1].\;$ From here, say, $(1-u)(1-v)\ge 0,\;$ implying $1+uv\ge u+v\;$ and, subsequently,

$1+w+uv\ge u+v+w.$

Therefore,

$\displaystyle \frac{1}{1+w+uv}\le \frac{1}{u+v+w}$

and

$\displaystyle \frac{w}{1+w+uv}\le \frac{w}{u+v+w}$

Similarly, $\displaystyle\frac{u}{1+u+vw}\le \displaystyle\frac{u}{u+v+w}\;$ and $\displaystyle\frac{v}{1+v+wu}\le \displaystyle\frac{v}{u+v+w}.$ Adding the three up we obtain

$\displaystyle\sum_{cycl}\frac{u}{1+u+vw}\le\sum_{cycl}\frac{u}{u+v+w}=1.$

Taking integral from $0\;$ to $1\;$ we obtain the required inequality. Equality is only possible for $a=b=c=0.$

### Acknowledgment

Dan Sitaru has kindly posted the above problem from his book Math Accent, at the CutTheKnotMath facebook page. He later communicated privately the solution above.