Question

Chứng minh:

\(\sqrt{2017}+\sqrt{2018}< \dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}}\)

Answer 1

Áp dụng bđt Svacxo ta có :

\(\dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}}\ge\dfrac{\left(\sqrt{2017}+\sqrt{2018}\right)^2}{\sqrt{2018}+\sqrt{2017}}=\sqrt{2017}+\sqrt{2018}\)

Dấu bằng xảy ra khi:

\(\dfrac{2017}{\sqrt{2018}}=\dfrac{2018}{\sqrt{2017}}\Leftrightarrow2017=2018\left(vl\right)\)

Suy ra không xảy ra dấu bằng

Vậy \(\dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}}>\sqrt{2017}+\sqrt{2018}\)

Answer 2

không thể cm