Quidditch wrote: Let $a, b, c \geq 1$. Prove that $$\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\geq\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ca}.$$ Let $a,b\geq1.$ Prove that $$\frac{1}{1+a^2}+\frac{1}{1+b^2}\geq \frac{2}{1+ab}$$

sqing wrote: Quidditch wrote: Let $a, b, c \geq 1$. Prove that $$\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\geq\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ca}.$$ For positive real numbers $a, b.$ Prove that $$\frac{1}{1+a^2}+\frac{1}{1+b^2}\geq \frac{2}{1+ab}$$ Assuming $a,b\ge 1$ as in the original problem statement: $\frac{1}{a^2+1} + \frac{1}{b^2+1}\ge \frac{2}{1+ab}$ $\iff a^3b + a^2 + ab + 1 + ab^3 + b^2 + ab + 1\ge 2a^2b^2 + 2a^2 +2 b^2 + 2$ $\iff a^3b + ab^3 + 2ab\ge 2a^2b^2 + a^2 + b^2$ $\iff a^3b + ab^3 - 2a^2b^2\ge a^2 + b^2 - 2ab$ $\iff ab(a-b)^2\ge (a-b)^2$ which follows from $a,b\ge 1$.

Thanks for the hint! Using the fact that $\frac{1}{1+a^2}+\frac{1}{1+b^2}\geq \frac{2}{1+ab},$ by symmetry we have $\frac{1}{1+a^2}+\frac{1}{1+c^2}\geq \frac{2}{1+ac}$ and $\frac{1}{1+b^2}+\frac{1}{1+c^2}\geq \frac{2}{1+bc}$. Adding the three terms, $\frac{1}{1+a^2}+\frac{1}{1+b^2} + \frac{1}{1+a^2}+\frac{1}{1+c^2} + \frac{1}{1+b^2}+\frac{1}{1+c^2} = 2 (\frac{1}{1+a^2}+\frac{1}{1+b^2} + \frac{1}{1+c^2}) \geq \frac{2}{1+ab} + \frac{2}{1+ac} + \frac{2}{1+bc}$. Hence $\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2} \geq \frac{1}{1+ab} + \frac{1}{1+ac} + \frac{1}{1+bc}$ as desired.

For any $x,y\ge 1$ and $k,m\in\mathbb{Z}_{>0}$ we have $\frac{k}{1+x^{k+m}}+\frac{m}{1+y^{k+m}}\ge\frac{k+m}{1+x^ky^m}$. Let's see a nice proof of this fact. Mine is hardly a nice one.