Question

[Lê Song Phương]

Cho \(x,y,z\) là 3 số thực dương thỏa mãn \(x^2+y^2+z^2=2\). Chứng minh: \(\dfrac{2}{x^2+y^2}+\dfrac{2}{y^2+z^2}+\dfrac{2}{z^2+x^2}\le\dfrac{x^3+y^3+z^3}{2xyz}+3\)

Answer 1

[nhạt]

Ta có: \(\dfrac{2}{x^2+y^2}=\dfrac{x^2+y^2+z^2}{x^2+y^2}=1+\dfrac{z^2}{x^2+y^2}\le1+\dfrac{z^2}{2xy}\)(bđt cosi)

CMTT: \(\dfrac{2}{y^2+z^2}\le1+\dfrac{x^2}{2yz}\); \(\dfrac{2}{z^2+x^2}\le1+\dfrac{y^2}{2xz}\)

=> \(\dfrac{2}{x^2+y^2}+\dfrac{2}{y^2+z^2}+\dfrac{2}{z^2+x^2}\le3+\dfrac{z^2}{2xy}+\dfrac{x^2}{2yz}+\dfrac{y^2}{2xz}=3+\dfrac{x^3+y^3+z^3}{2xyz}\) (Đpcm)

Answer 2

[Nguyễn Trần Gia Bảo]

Ta có: $\frac{2}{x^2+y^2}=\frac{x^2+y^2+z^2}{x^2+y^2}=1+\frac{z^2}{x^2+y^2}\le 1+\frac{z^2}{2xy}$(bđt cosi)

CMTT: $\frac{2}{y^2+z^2}\le 1+\frac{x^2}{2yz}$; $\frac{2}{z^2+x^2}\le 1+\frac{y^2}{2xz}$

=> $\frac{2}{x^2+y^2}+\frac{2}{y^2+z^2}+\frac{2}{z^2+x^2}\le 3+\frac{z^2}{2xy}+\frac{x^2}{2yz}+\frac{y^2}{2xz}=3+\frac{x^3+y^3+z^3}{2xyz}$ ( Đpcm )