Question

Tính \(\sqrt{2016}+\sqrt{2018}\) và \(2\sqrt{2017}\)

Answer 1

Áp dụng bất đẳng thức Bu - nhi - a

\(\left(1\cdot\sqrt{2016}+1\cdot\sqrt{2018}\right)^2\le\left(1^2+1^2\right)\cdot\left[\left(\sqrt{2016}\right)^2+\left(\sqrt{2018}\right)^2\right]=2\cdot4034\)

=8068

\(\Rightarrow\sqrt{2016}+\sqrt{2018}\le\sqrt{8068}=\sqrt{4\cdot2017}=2\sqrt{2017}\)

\(\)Vậy \(\sqrt{2016}+\sqrt{2018}\le2\sqrt{2017}\)