Question

Chứng minh 

\(\sqrt{\frac{7}{2}}\le\left|1+z\right|+\left|1-z+z^2\right|\le\sqrt[3]{\frac{7}{6}}\), với mọi \(z,\left|z\right|=1\)

Answer 1

Đặt \(t=\left|1+z\right|\in\left[0,2\right]\) 

\(t^2=\left(1+z\right)\left(1+\overline{z}\right)=2+2Re\left(z\right)\) 

\(\Rightarrow Re\left(z\right)=\frac{t^2-2}{2}\)

Khi đó \(\left|1-z+z^2\right|=\sqrt{\left|7-2t^2\right|}\)

Xét hàm số :

\(f:\left[0;2\right]\) -> \(R,f\left(t\right)=t+\sqrt{\left|7-2t^2\right|}\)

Ta được :

\(f\left(\sqrt{\frac{7}{2}}\right)=\sqrt{\frac{7}{2}}\le t+\sqrt{\left|7-2t^2\right|}\le f\left(\sqrt{\frac{7}{2}}\right)=\sqrt[3]{\frac{7}{6}}\)