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Bài 1:Áp dụng C-S dạng engel
\(\frac{3}{xy+yz+xz}+\frac{2}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{2}{x^2+y^2+z^2}\)
\(\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{2}\right)^2>14\)
Bổ đề: xyz+(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)
Cm:
VT: xyz+(x+y)(y+z)(z+x)
=xyz+xyz+x2z+x2y+y2x+y2z+z2x+z2y+xyz
=xyz+x2z+x2y+xyz+y2z+y2x+xyz+z2x+z2y
=(x+y+z)(xy+yz+xz)
AD bổ đề và đề bài cho
=> (x+y)(y+z)(z+x)=0
\(\Rightarrow\left[{}\begin{matrix}x+y=0\\x+z=0\\y+z=0\end{matrix}\right.\)
1. x+y=0
ta có x2019+y2019=(x+y)(x2018-x2017y+...+y2018)=0
=> x2019+y2019+z2019=z2019
Có (x+y+z)2019=z2019
=> x2019+y2019+z2019= (x+y+z)2019
Làm tương tự với 2 trường hợp còn lại ta được đpcm
Từ gt suy ra: \(x+\sqrt{x^2+2019}=\dfrac{2019}{y+\sqrt{y^2+2019}}=\sqrt{y^2+2019}-y\).
Tương tự: \(y+\sqrt{y^2+2019}=\sqrt{x^2+2019}-x\).
Do đó dễ dàng suy ra được: \(x+y=0\).
\(\Rightarrow x=-y\Rightarrow x^{2019}+y^{2019}=x^{2019}+\left(-x\right)^{2019}=0\left(đpcm\right)\).
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{z}{\left(x+y+z\right).z}-\frac{x+y+z}{z.\left(x+y+z\right)}=\frac{-x-y}{z.\left(x+y+z\right)}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{x+y}{-z.\left(x+y+z\right)}\)
TH1: x+y=0
=> x=-y => P=0
TH2: xy=-z.(x+y+z)
\(\Leftrightarrow xy=-xz-zy-z^2\Leftrightarrow xy+xz+zy+z^2=0\Leftrightarrow x.\left(y+z\right)+z.\left(y+z\right)=0\)
\(\Leftrightarrow\left(x+z\right).\left(y+z\right)=0\Leftrightarrow\orbr{\begin{cases}x=-z\\y=-z\end{cases}\Rightarrow P=0}\)
Ta có: x2 + y2 + z2 = xy + yz + zx
<=> [(x - y)2 + (y - z)2 + (z - x)2] . 1/2 = 0
<=> x = y = z
Thay vào pt thứ 2...
Có \(y^2+2019=y^2+xy+yz+zx=y\left(x+y\right)+z\left(x+y\right)=\left(y+z\right)\left(x+y\right)\)
\(x^2+2019=x^2+xy+yz+zx=x\left(x+y\right)+z\left(x+y\right)=\left(x+z\right)\left(x+y\right)\)
\(z^2+2019=z^2+xy+yz+xz=z\left(z+y\right)+x\left(y+z\right)=\left(z+x\right)\left(y+z\right)\)
Có \(P=x\sqrt{\frac{\left(y^2+2019\right)\left(z^2+2019\right)}{x^2+2019}}+y\sqrt{\frac{\left(z^2+2019\right)\left(x^2+2019\right)}{y^2+2019}}+z\sqrt{\frac{\left(x^2+2019\right)\left(y^2+2019\right)}{z^2+2019}}\)
=\(x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(z+y\right)}{\left(x+z\right)\left(y+x\right)}}+y\sqrt{\frac{\left(z+x\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(y+z\right)\left(x+y\right)}}+z\sqrt{\frac{\left(x+z\right)\left(x+y\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)
=\(x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
=\(x\left|y+z\right|+y\left|x+z\right|+z\left|x+y\right|\)
=\(x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\) (vì x,y,z >0)
= xy+xz+xy+yz+xz+yz
=2(xy+xz+yz)=2.2019(vì xy+xz+yz=2019)
=4038
Vậy P=4038