Lời giải:
Ta thấy, với mọi $x,y,z$ là số thực thì:

$(x-y+z)^2\geq 0$

$\sqrt{y^4}\geq 0$

$|1-z^3|\geq 0$

$\Rightarrow (x-y+z)^2+\sqrt{y^4}+|1-z^3|\geq 0$ với mọi $x,y,z$

Kết hợp $(x-y+z)^2+\sqrt{y^4}+|1-z^3|\leq 0$

$\Rightarrow (x-y+z)^2+\sqrt{y^4}+|1-z^3|=0$

Điều này xảy ra khi: $x-y+z=y^4=1-z^3=0$

$\Leftrightarrow y=0; z=1; x=-1$

x+y>=2 căn xy

y+z>=2 căn yz

x+z>=2 căn xz

=>(x+y)(y+z)(x+z)>=8xyz

lllllllllllllllllllllllllllllllllllllllllllllllllllllll

mn giúp mình với

\(\left(x-y\right)^2\ge0\)

\(\Rightarrow\left(x+y\right)^2\ge4xy\)

\(\Rightarrow xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{2019^2}{4}\)

Dấu = xảy ra khi \(x=y=\dfrac{2019}{2}\)

\(P_{max}=1019090\)

\(xy+2yz+3zx=xy+zx+2yz+2zx=x\left(y+z\right)+2z\left(y+x\right)=x.\left(-x\right)+2z.\left(-z\right)=-x^2-2z^2\le0\)-Dấu bằng xảy ra \(\Leftrightarrow x=y=z=0\)

\(\dfrac{x^2+y^2}{a^2+b^2}=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\)

\(\Leftrightarrow\dfrac{x^2+y^2}{a^2+b^2}=\dfrac{x^2b^2+a^2y^2}{a^2b^2}\)

\(\Leftrightarrow\left(x^2+y^2\right)a^2b^2=\left(a^2+b^2\right)\left(x^2b^2+a^2y^2\right)\)

\(\Leftrightarrow a^2b^2x^2+a^2b^2y^2=a^2x^2b^2+a^4y^2+b^4x^2+a^2y^2b^2\)

\(\Leftrightarrow0=a^4y^2+b^4x^2\)

Có \(\left\{{}\begin{matrix}a^4y^2\ge0\\b^4x^2\ge0\end{matrix}\right.\) =>\(a^4y^2+b^4x^2\ge0\)

 [=] xảy ra <=> \(\left\{{}\begin{matrix}a^4y^2=0\\b^4x^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\) (vì a;b khác 0)

Vậy y=x=0 (đpcm)