Question

Cho x, y thuộc R. C/m:

\(\frac{1}{\left(1+x\right)^2}+\frac{1}{\left(1+y\right)^2}\ge\frac{1}{1+xy}\)

Answer 1

Ta có x, y ∈ R nên \(\frac{1}{\left(1+x\right)^2}+\frac{1}{\left(1+y\right)^2}\ge\frac{1}{1+xy}\)

\(\Leftrightarrow\left(1+xy\right)\left[\left(1+x\right)^2+\left(1+y\right)^2\right]\ge\left(1+x\right)^2\left(1+y\right)^2\\ \Leftrightarrow\left(1+xy\right)\left[2\left(1+x+y\right)+x^2+y^2\right]\ge\left[\left(1+x+y\right)+xy\right]^2\\ \Leftrightarrow1+xy\left(x^2+y^2\right)\ge2xy+x^2y^2\\ \Leftrightarrow xy\left(x^2+y^2-2xy\right)+\left(x^2y^2-2xy+1\right)\ge0\\ \Leftrightarrow xy\left(x-y\right)^2+\left(xy-1\right)^2\ge0\)

đúng vì x, y > 0