Ta có: P2 = \(\left(2015x+2016\sqrt{1-x^2}\right)^2\)
Áp dụng bất đẳng thức Bunhiacopxki ta có:
\(\left(2015x+2016\sqrt{1-x^2}\right)^2\) ≤ ( 20152 + 20162 )( x2 + 1 - x2 ) = 20152 + 20162
=> P ≤ \(\sqrt{2015^2+2016^2}\)
Dấu " = " xảy ra khi \(\frac{x}{2015}=\frac{2016}{\sqrt{1-x^2}}=>x=\frac{2015}{\sqrt{2015^2+2016^2}}\)