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\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)=> \(\frac{xy+yz+zx}{xyz}=\frac{1}{x+y+z}\)
=> (x+y+z)(xy+yz+zx) = xyz
=> \(x^2y+xy^2+y^2z+yz^2+zx^2+z^2x+2xyz=0\)
=> (x+y)(y+z)(z+x) = 0
=> \(\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
TH1: x = -y
=> \(\frac{1}{x^{2019}}+\frac{1}{y^{2019}}+\frac{1}{z^{2019}}=\frac{1}{\left(-y\right)^{2019}}+\frac{1}{y^{2019}}+\frac{1}{z^{2019}}=\frac{1}{z^{2019}}\)
=> \(\frac{1}{x^{2019}+y^{2019}+z^{2019}}=\frac{1}{\left(-y\right)^{2019}+y^{2019}+z^{2019}}=\frac{1}{z^{2019}}\)
=> ĐPCM
Tương tự với TH2 và TH3
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\cdot\frac{xy+z\left(x+y+z\right)}{xyz\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow x=-y\left(h\right)y=-z\left(h\right)z=-x\)
Nếu
\(x=-y\Rightarrow\frac{1}{x^{2019}}+\frac{1}{y^{2019}}+\frac{1}{z^{2019}}=\frac{1}{x^{2019}}-\frac{1}{x^{2019}}+\frac{1}{z^{2019}}=\frac{1}{z^{2019}}\)
\(\frac{1}{x^{2019}+y^{2019}+z^{2019}}=\frac{1}{x^{2019}-x^{2019}+z^{2019}}=\frac{1}{z^{2019}}\)
Tương tự các TH còn lại nha!
P/S:Có 1 bài chặt hơn ntnày:
Cho \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\) thì \(\frac{1}{x^n}+\frac{1}{y^n}+\frac{1}{z^n}=\frac{1}{x^n+y^n+z^n}\) với n lẻ.
Sửa đề : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2019}\)
Thay \(2019=x+y+z\)ta có :
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}\)
\(\Leftrightarrow\frac{y}{xy}+\frac{x}{xy}=\frac{z}{z\left(x+y+z\right)}-\frac{x+y+z}{z\left(x+y+z\right)}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{z-x-y-z}{z\left(x+y+z\right)}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{-\left(x+y\right)}{z\left(x+y+z\right)}\)
\(\Leftrightarrow z\left(x+y\right)\left(x+y+z\right)=-xy\left(x+y\right)\)
\(\Leftrightarrow z\left(x+y\right)\left(x+y+z\right)+xy\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left[z\left(x+y+z\right)+xy\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left(xz+yz+z^2+xy\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left[z\left(x+z\right)+y\left(x+z\right)\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left(x+z\right)\left(y+z\right)=0\)
( mình chỉ xét 1 t/h, các t/h còn lại hoàn toàn tương tự )
TH1 : \(x+y=0\)
\(\Leftrightarrow x=-y\)(1)
Thay (1) vào A ta có :
\(A=\frac{1}{-y^{2019}}+\frac{1}{y^{2019}}+\frac{1}{z^{2019}}\)
\(A=\frac{1}{z^{2019}}\)
Mặt khác : \(x+y+z=2019\)
Thay (1) vào đẳng thức trên ta được : \(-y+y+z=2019\)
\(\Leftrightarrow z=2019\)
Thay z vào A ta được : \(A=\frac{1}{2019^{2019}}\)
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\Rightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\)
\(\Rightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)\(\Rightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Rightarrow\left(x+y\right)\left(\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)\(\Rightarrow\left(x+y\right)\left[z\left(x+z\right)+y\left(x+z\right)\right]=0\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)\(\Rightarrow\)\(x=-y\) hoặc \(y=-z\) hoặc \(z=-x\)
\(\Rightarrow A=0\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{2x+y+z}\leq \frac{1}{16}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\frac{1}{x+2y+z}\leq \frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\frac{1}{x+y+2z}\leq \frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}\right)\)
Cộng theo vế 3 BĐT trên thu được:
\(M\leq \frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{2019}{4}\)
Vậy $M_{\max}=\frac{2019}{4}$. Giá trị này đạt tại $x=y=z=\frac{3}{2019}$
