Sửa đề : Cho \(x;y\text{ ≥}1\) . CMR : \(\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}\) ≥ \(\dfrac{2}{xy+1}\)
---- Lời giải ----
\(\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}\) ≥ \(\dfrac{2}{xy+1}\)
⇔ \(\dfrac{1}{x^2+1}-\dfrac{1}{xy+1}+\dfrac{1}{y^2+1}-\dfrac{1}{xy+1}\text{≥}0\)
⇔\(\dfrac{x\left(y-x\right)}{\left(x^2+1\right)\left(xy+1\right)}+\dfrac{y\left(x-y\right)}{\left(y^2+1\right)\left(xy+1\right)}\text{≥}0\)
⇔ \(\dfrac{y\left(x-y\right)}{\left(y^2+1\right)\left(xy+1\right)}\text{≥}\dfrac{x\left(x-y\right)}{\left(x^2+1\right)\left(xy+1\right)}\)
⇔ \(\dfrac{y}{y^2+1}\text{≥}\dfrac{x}{x^2+1}\)
⇔ \(y\left(x^2+1\right)\text{≥}x\left(y^2+1\right)\)
⇔ \(x^2y-xy^2+y-x\text{≥}0\)
⇔\(\left(x-y\right)\left(xy-1\right)\text{≥}0\) ( luôn đúng vì : xy ≥ 1 )
KL..............
