Question

Cho các số x, y thoả mãn x+y\(\ne0\), chứng minh:

x² + y² + \(\left(\dfrac{1+xy}{x+y}\right)^{^2}\)\(\ge2\)

Answer 1

Theo mình nó còn có x,y > 0 nữa nha !

Ta có:

\(x^2+y^2+\left(\dfrac{1+xy}{x+y}\right)^2=\left(x+y\right)^2+\left(\dfrac{1+xy}{x+y}\right)^2-2xy\)

Áp dụng BĐT Cosi ta có:

\(\left(x+y\right)^2+\left(\dfrac{1+xy}{x+y}\right)^2\ge2\sqrt{\left(x+y\right)^2\left(\dfrac{1+xy}{x+y}\right)^2}=2\left(1+xy\right)\)

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

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

\(\Rightarrow\)đpcm