Áp dụng hệ quả BĐT thức Côsi với 3 số : (a+b+c )³ > 27abc ta có: xyz < = (x+y+z)³/27 ; (x+y)(y+z)(z+x) < = ( x+y+z+x+y+z)/27

==> M < = (2³/27)(4³/27)=512/729

==> M max = 512/729 <=> x = y = z = 2/3

\(\dfrac{xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\le\dfrac{xyz}{2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}}=\dfrac{1}{8}\)

Dấu "=" xảy ra khi \(x=y=z\)

\(A=\left(x+y-z\right)\left(y+z-x\right)\left(z+x-y\right)\)

\(áp\) \(dụng\) \(bđt:\) \(\)\(AM-GM:a+b\ge2\sqrt{ab}\Leftrightarrow\sqrt{ab}\le\dfrac{a+b}{2}\Leftrightarrow ab\le\dfrac{\left(a+b\right)^2}{4}\)

\(\Rightarrow A^2=\left(x+y-z\right)^2\left(y+z-x\right)^2\left(z+x-y^2\right)=\left(x+y-z\right)\left(y+z-x\right)\left(y+z-x\right)\left(z+x-y\right)\left(x+y-z\right)\left(z+x-y\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x+y-z\right)\left(y+z-x\right)\le\dfrac{\left(x+y-z+y+z-x\right)^2}{4}\le\dfrac{4y^2}{4}\le y^2\\\left(y+z-x\right)\left(z+x-y\right)\le\dfrac{\left(y+z-x+z+x-y\right)^2}{4}\le z^2\\\left(x+y-z\right)\left(z+x-y\right)\le\dfrac{\left(x+y-z+z+x-y\right)^2}{4}\le x^2\\\end{matrix}\right.\)

\(\)\(\Rightarrow A^2\le x^2y^2z^2\le\left(xyz\right)^2\Rightarrow A\le xyz\)