一非常硬 (one very hard)

tommywong

teomihai:

soruz
1056 posts
#1VPMYesterday at 11:04 PM
If $a \ge b\ge c\ge d >0$ such that $ab+bc+cd+da=4$, then
$$\frac 1{ab+4}+\frac 1{ac+4}+\frac 1{ad+4}+\frac 1{bc+4}+\frac 1{bd+4}+\frac 1{cd+4}\ge \frac{6}{5} .$$

teomihai:

Maybe help this:
Let $a,b,c,d$ to be non negative real numbers satisfying $ab+ac+ad+bc+bd+cd=6$. Prove that

\[\dfrac{1}{a^2+1} + \dfrac{1}{b^2+1} + \dfrac{1}{c^2+1} + \dfrac{1}{d^2+1} \ge 2\]?
thanks.