Question

Cho 2 số x,y khác 0. Chứng minh rằng:

\(x^2+y^2+\left(\frac{1+xy}{x+y}\right)^2\ge2\)

Answer 1

\(x^2+y^2+\left(\frac{1+xy}{x+y}\right)^2\ge2\)

\(\Leftrightarrow\left(x+y\right)^2-2xy+\left(\frac{1+xy}{x+y}\right)^2\ge2\)

\(\Leftrightarrow\left(x+y\right)^2-2\left(xy+1\right)+\left(\frac{1+xy}{x+y}\right)^2\ge0\)

\(\Leftrightarrow\left(x+y\right)^2-\frac{2\left(x+y\right)\left(xy+1\right)}{\left(x+y\right)}+\left(\frac{1+xy}{x+y}\right)^2\ge0\)

\(\Leftrightarrow\left(x+y-\frac{xy+1}{x+y}\right)^2\ge0\) (đúng)

Vậy ...