tìm x1,x2,x3,...,xn thỏa mãn căn x^2-1^2

Áp dụng bất đẳng thức Cosi:

\(1.\sqrt {{x_1}^2 - {1^2}} \le \frac{{{1^2} + {x_1}^2 - {1^2}}}{2} = \frac{{{x_1}^2}}{2}\)

\(2.\sqrt {{x_2}^2 - {2^2}} \le \frac{{{2^2} + {x_2}^2 - {2^2}}}{2} = \frac{{{x_2}^2}}{2}\)

….

\(n.\sqrt {{x_n}^2 - {n^2}} \le \frac{{{n^2} + {x_n}^2 - {n^2}}}{2} = \frac{{{x_n}^2}}{2}\)

Cộng theo vế ta được:

\(\sqrt {{x_1}^2 - {1^2}} + 2\sqrt {{x_2}^2 - {2^2}} + ... + n\sqrt {{x_n}^2 - {n^2}} \le \frac{{{x_1}^2}}{2} + \frac{{{x_2}^2}}{2} + ... + \frac{{{x_n}^2}}{2}\)

\(\sqrt {{x_1}^2 - {1^2}} + 2\sqrt {{x_2}^2 - {2^2}} + ... + n\sqrt {{x_n}^2 - {n^2}} \le \frac{1}{2}\left( {{x_1}^2 + {x_2}^2 + ... + {x_n}^2} \right)\)

Dấu “=” xảy ra khi: \(\left\{ \begin{array}{l}\sqrt {{x_1}^2 - {1^2}} = 1\\\sqrt {{x_2}^2 - {2^2}} = 2\\....\\\sqrt {{x_n}^2 - {n^2}} = n\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x_1} = \sqrt 2 \\{x_2} = 2\sqrt 2 \\....\\{x_n} = n\sqrt 2 \end{array} \right.\)