Cách 1:

Ta có \(A=xy+yz+2zx\)

\(\Rightarrow A+1=x^2+y^2+z^2+xy+yz+2zx\)

                    \(=\left(x+z+\frac{y}{2}\right)^2+\frac{3}{4}y^2\ge0\)

\(\Rightarrow A\ge-1\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}y=0\\x=-z\end{cases}}\)

Ta có : \(\left(x+y+z\right)^2\ge0\)

\(\Rightarrow xy+yz+zx\ge\frac{-\left(x^2+y^2+z^2\right)}{2}=-\frac{1}{2}\)

Lại có : \(\left(x+z\right)^2\ge0\Rightarrow xz\ge\frac{-\left(x^2+z^2\right)}{2}=\frac{y^2-1}{2}\ge-\frac{1}{2}\)

Khi đó : \(xy+yz+2zx\ge-1\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}y=o\\x^2=z^2=\frac{1}{2}\end{cases}}\)

Cách 2

Ta có : \(\left(x+y+z\right)^2\ge0\)

\(\Leftrightarrow xy+yz+zx\ge\frac{-\left(x^2+y^2+z^2\right)}{2}=\frac{-1}{2}\)

Mặt khác: \(\left(x+z\right)^2\ge0\Rightarrow xz\ge\frac{-\left(x^2+z^2\right)}{2}=\frac{y^2-1}{2}\ge\frac{-1}{2}\)

\(\Rightarrow xy+yz+2xz\ge-1\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}y=0\\x=-z\end{cases}}\)

khó quá!!!!!!!!!!!

Đáp án là -13 bn ơi

Áp dụng BĐT (a - b)² ≥ 0 → a² + b² ≥ 2ab ta có: 

+) x² + y² ≥ 2xy 

x² + 1 ≥ 2x 

+) y² + z² ≥ 2yz 

y² + 1 ≥ 2y 

+) z² + x² ≥ 2xz 

z² + 1 ≥ 2z 

=> 2 ( x² + y² + z² ) ≥ 2( xy + yz + xz )
cộng các BĐT trên ta có
3( x² + y² + z² ) + 3 ≥ 2( x + y + z + xy + yz + xz)
=> GTNN của P = 3 khi và chỉ khi x=y=z=1

Nó bảo sai bn ơi

\(T\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+x+y+z}=\dfrac{x+y+z}{2}\ge\dfrac{2019}{2}\)

áp dụng BĐT:\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\) với a,b,c,x,y,z là số dương

ta có BĐT Bunhiacopxki cho 3 bộ số:\(\left(\dfrac{a}{\sqrt{x}};\sqrt{x}\right);\left(\dfrac{b}{\sqrt{y}};\sqrt{y}\right);\left(\dfrac{c}{\sqrt{z}};\sqrt{z}\right)\)

ta có :

\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\left(x+y+z\right)\)\(=\left[\left(\dfrac{a}{\sqrt{x}}\right)^2+\left(\dfrac{b}{\sqrt{y}}\right)^2+\left(\dfrac{c}{\sqrt{z}}\right)^2\right]\).\(\left[\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2+\left(\sqrt{z}\right)^2\right]\)\(\ge\left(\dfrac{a}{\sqrt{x}}.\sqrt{x}+\dfrac{b}{\sqrt{y}}.\sqrt{y}+\dfrac{c}{\sqrt{z}}.\sqrt{z}\right)^2=\left(a+b+c\right)^2\)

lúc đó ta có :\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\ge\dfrac{\left(a+b+c\right)^2}{x+y+z}\)

ta có \(T=\dfrac{x^2}{x+\sqrt{yz}}+\dfrac{y^2}{y+\sqrt{zx}}+\dfrac{z^2}{z+\sqrt{xy}}\)\(\ge\dfrac{\left(x+y+z\right)^2}{x+\sqrt{yz}+y+\sqrt{zx}+z+\sqrt{xy}}\) mà ta có :

\(\sqrt{yz}+\sqrt{zx}+\sqrt{xy}\)\(\le\dfrac{x+y}{2}+\dfrac{x+z}{2}+\dfrac{z+y}{2}\)\(\Rightarrow\sqrt{yz}+\sqrt{zx}+\sqrt{xy}\le x+y+z\)

\(\Rightarrow T=\dfrac{2019}{2}\Leftrightarrow x=y=z=673\)

vậy \(\text{MinT}=\dfrac{2019}{2}\) khi và chỉ khi x=y=z=673

b,Ap dung bdt cauchy schwarz dang engel ta co

\(B=\frac{x^2}{1}+\frac{y^2}{1}+\frac{z^2}{1}>=\frac{\left(x+y+z\right)^2}{3}=\frac{a^2}{3}\)

xay ra dau = khi x=y=z=a/3

Ta có :

\(x^2+y^2\ge2xy\)

\(y^2+z^2\ge2yz\)

\(z^2+x^2\ge2zx\)

\(x^2+1\ge2x\)

\(y^2+1\ge2y\)

\(z^2+1\ge2z\)

Suy ra :  \(3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+zx\right)\)

\(\Leftrightarrow3\left(x^2+y^2+z^2\right)+3\ge2.6=12\)

\(\Leftrightarrow3\left(x^2+y^2+z^2\right)\ge9\)

\(\Leftrightarrow x^2+y^2+z^2\ge3\)

Dấu ''='' xảy ra khi x=y=z=1

Vậy GTNN của  \(x^2+y^2+z^2\)là 3 khi x=y=z=1