Question

a) Chứng minh: \(2016^{2015}+2018^{2016}⋮2017\)

b) Cho x, y \(\ge\)1

Chứng minh: \(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}\ge\dfrac{2}{1+xy}\)

Answer 1

b) \(x,y\ge1\Rightarrow xy\ge1\)

BĐT đã cho tương đương với:

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

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

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

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

BĐT cuối luôn đúng nên ta có đpcm

Đẳng thức xảy ra khi x=y hoặc xy=1