Ta có : \(\left(2x-1\right)^{2014}\) \( \geq\) \(0\) với mọi x
\(\left(x-y-\frac{1}{2}\right)^{2016}\) \( \geq\) \(0\) với mọi x ; y
\(\implies\)\(\left(2x-1\right)^{2014}+\left(x-y-\frac{1}{2}\right)^{2016}\) \( \geq\) \(0\) với mọi x ; y
Mà \(\left(2x-1\right)^{2014}+\left(x-y-\frac{1}{2}\right)^{2016}=0\)
\(\implies\) Dấu bằng xảy ra \(\iff\) \(\hept{\begin{cases}\left(2x-1\right)^{2014}=0\\\left(x-y-\frac{1}{2}\right)^{2016}=0\end{cases}}\) \(\iff\) \(\hept{\begin{cases}2x-1=0\\x-y-\frac{1}{2}=0\end{cases}}\) \(\iff\) \(\hept{\begin{cases}x=\frac{1}{2}\\y=0\end{cases}}\)
