Problem - 3227
Let $a, b, c$ be positive numbers lying in the interval $(0, 1]$. Show that $$\frac{a}{1+b+ca}+\frac{b}{1+c+ab}+\frac{c}{1+a+bc}\le 1$$
Because $$1+ab=(1-a)(1-b)+a+b$$
We have $$1+c+ab=(1-a)(1-b)+a+b+c\ge a+b+c$$
Therefore $$\frac{a}{1+b+ca}+\frac{b}{1+c+ab}+\frac{c}{1+a+bc}\le\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1$$