Question

Cho x,y,z > 0 , x + y + z <= \(\frac{3}{2}\). C/m : \(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}>=\frac{3}{2}\sqrt{17}\)

Answer 1

\(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)

\(\ge\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}=\sqrt{\left(x+y+z\right)^2+\frac{81}{16\left(x+y+z\right)^2}+\frac{1215}{16\left(x+y+z\right)^2}}\)

\(\ge\sqrt{2\sqrt{\frac{81\left(x+y+z\right)^2}{16\left(x+y+z\right)^2}}+\frac{1215}{16.\left(\frac{3}{2}\right)^2}}=\frac{3\sqrt{17}}{2}\)

Dấu "=" xảy ra khi \(z=y=z=\frac{1}{2}\)