Anh giải thử không biết đúng koDK $-1\leq x_1,x_2,x_3,...,x_{2000}\leq 1$đặt $a=x_1+x_2+..+x_{2000}$$(1) VT=2000\sqrt{\frac{2001}{2000}}\leq \sqrt{2000(2000+a)}\Rightarrow a\geq 1$ ( bunhia)$(2)VT=2000\sqrt{\frac{1999}{2000}}\leq \sqrt{2000(2000-a)}\Rightarrow a\leq 1$ ( bunhia)$\Rightarrow a=1\Rightarrow x_1=x_2=...=x_{2000}$Vậy $x_1=x_2=x_3=...=x_{2000}=\frac{1}{2000}$
Anh giải thử không biết đúng koDK $-1\leq x_1,x_2,x_3,...,x_{2000}\leq 1$đặt $a=x_1+x_2+..+x_{2000}$$(1) VT=2000\sqrt{\frac{2001}{2000}}\leq \sqrt{2000(2000+a)}\Rightarrow a\geq 1$ ( bunhia)$(2)VT=2000\sqrt{\frac{1999}{2000}}\leq \sqrt{2000(2000-a)}\Rightarrow a\leq 1$ ( bunhia)$\Rightarrow a=1\Rightarrow x_1=x_2=...=x_{2000}$$x_1=x_2=x_3=...=x_{2000}=\frac{1}{2000}$
Anh giải thử không biết đúng koDK $-1\leq x_1,x_2,x_3,...,x_{2000}\leq 1$đặt $a=x_1+x_2+..+x_{2000}$$(1) VT=2000\sqrt{\frac{2001}{2000}}\leq \sqrt{2000(2000+a)}\Rightarrow a\geq 1$ ( bunhia)$(2)VT=2000\sqrt{\frac{1999}{2000}}\leq \sqrt{2000(2000-a)}\Rightarrow a\leq 1$ ( bunhia)$\Rightarrow a=1\Rightarrow x_1=x_2=...=x_{2000}$
Vậy $x_1=x_2=x_3=...=x_{2000}=\frac{1}{2000}$