Question

Cho các số thực dương x,y. CMR: \(\dfrac{1}{\left(1+x\right)^2}+\dfrac{1}{\left(1+y\right)^2}\ge\dfrac{1}{1+xy}\)

Answer 1

\(\left(1+x\right)^2=\left(1.1+\sqrt{xy}.\sqrt{\dfrac{x}{y}}\right)^2\le\left(1+xy\right)\left(1+\dfrac{x}{y}\right)=\dfrac{\left(1+xy\right)\left(x+y\right)}{y}\)

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

Tương tự ta có: \(\dfrac{1}{\left(1+y\right)^2}\ge\dfrac{x}{\left(1+xy\right)\left(x+y\right)}\)

Cộng vế với vế: 

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

Dấu "=" xảy ra khi \(x=y=1\)