Question

Cho x,y,z là các số thực dương

CMR: \(\dfrac{1}{x^2+yz}+\dfrac{1}{y^2+zx}+\dfrac{1}{z^2+xy}\le\dfrac{x+y+z}{2xyz}\)

Answer 1

Ta có:

\(VT=\dfrac{1}{x^2+yz}+\dfrac{1}{y^2+xz}+\dfrac{1}{z^2+xy}\le\dfrac{1}{2x\sqrt{yz}}+\dfrac{1}{2y\sqrt{xz}}+\dfrac{1}{2z\sqrt{xy}}\)

\(\Rightarrow VT\le\dfrac{\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}{2xyz}\le\dfrac{\dfrac{x+y}{2}+\dfrac{y+z}{2}+\dfrac{x+z}{2}}{2xyz}=\dfrac{x+y+z}{2xyz}\)

Dấu "=" xảy ra khi \(x=y=z\)