cho x+y+z =1 min $\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}$ta có theo bdt mincopxiki
$\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}>=\sqrt{(x+y+z)^2+(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2}>=\sqrt{82}$
vì ta có bdt svac dc $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}>=9$
dấu bằng xảy ra khi x=y=z=1/3