Question

Xác định dạng của ΔABC biết

\(\sqrt[n]{sinA}+\sqrt[n]{sinB}=2\sqrt[n]{cos\dfrac{C}{2}}\)

Answer 1

Sử dụng BĐT: \(a^n+b^n\ge2\left(\dfrac{a+b}{2}\right)^n\) hay \(\left(a+b\right)^n\le2^{n-1}\left(a^n+b^n\right)\)

Ta có: 

\(2^ncos\dfrac{C}{2}=\left(\sqrt[n]{sinA}+\sqrt[n]{sinB}\right)^n\le2^{n-1}\left(sinA+sinB\right)\)

\(\Leftrightarrow2cos\dfrac{C}{2}\le2sin\dfrac{A+B}{2}cos\dfrac{A-B}{2}\)

\(\Leftrightarrow cos\dfrac{C}{2}\le cos\dfrac{C}{2}.cos\dfrac{A-B}{2}\)

\(\Leftrightarrow cos\dfrac{A-B}{2}=1\Leftrightarrow A=B\)

Tam giác cân tại C