Question

Answer 1

Áp dụng BĐT cosi: \(a+b\ge2\sqrt{ab}\Leftrightarrow ab\le\dfrac{\left(a+b\right)^2}{4}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y-z\right)\left(y+z-x\right)\le\dfrac{\left(2y\right)^2}{4}=y^2\\\left(y+z-x\right)\left(z+x-y\right)\le\dfrac{\left(2z\right)^2}{4}=z^2\\\left(y+z-x\right)\left(z+x-y\right)\le\dfrac{\left(2x\right)^2}{4}=x^2\end{matrix}\right.\\ \Leftrightarrow\left[\left(x+y-z\right)\left(y+z-x\right)\left(z+x-y\right)\right]^2\le\left(xyz\right)^2\\ \Leftrightarrow\left(x+y-z\right)\left(y+z-x\right)\left(z+x-y\right)\le xyz\)

Dấu \("="\Leftrightarrow x=y=z\)