Question

1,cho 3 số thực dương x,y,z thỏa mãn x+y≤z.CMR:\(\left(x^2+y^2+z^2\right)\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\ge\frac{27}{2}\)

(giải giúp bằng bất phương trình)

Answer 1

\(x+y\le z\Rightarrow\frac{z}{x+y}\ge1\)\(VT=3+\frac{x^2}{y^2}+\frac{y^2}{x^2}+\frac{x^2}{z^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}+\frac{z^2}{y^2}\)

\(VT=3+\left(\frac{x^2}{y^2}+\frac{y^2}{x^2}\right)+\left(\frac{x^2}{z^2}+\frac{z^2}{16x^2}\right)+\left(\frac{y^2}{z^2}+\frac{z^2}{16y^2}\right)+\frac{15z^2}{16}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\)

\(VT\ge3+2\sqrt{\frac{x^2y^2}{x^2y^2}}+2\sqrt{\frac{x^2z^2}{16x^2z^2}}+2\sqrt{\frac{y^2z^2}{16y^2z^2}}+\frac{15z^2}{32}\left(\frac{1}{x}+\frac{1}{y}\right)^2\)

\(VT\ge3+2+\frac{1}{2}+\frac{1}{2}+\frac{15z^2}{32}\left(\frac{4}{x+y}\right)^2\)

\(VT\ge6+\frac{15}{2}\left(\frac{z}{x+y}\right)^2\ge6+\frac{15}{2}=\frac{27}{2}\)

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