\(\dfrac{x}{2018}=\dfrac{y}{2019}=\dfrac{x-y}{-1};\dfrac{y}{2019}=\dfrac{z}{2020}=\dfrac{y-z}{-1};\dfrac{x}{2018}=\dfrac{z}{2020}=\dfrac{x-z}{-2}\\ \Leftrightarrow\dfrac{x-y}{-1}=\dfrac{y-z}{-1}=\dfrac{x-z}{-2}\\ \Leftrightarrow2\left(x-y\right)=2\left(y-z\right)=x-z\\ \Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^3=8\left(x-y\right)^2\left(x-y\right)=8\left(x-y\right)^2\left(y-z\right)\)

Lời giải:

Đặt $\frac{x}{2018}=\frac{y}{2019}=\frac{z}{2020}=a$

$\Rightarrow x=2018a; y=2019a; z=2020a$

$\Rightarrow (x-z)^3=(2018a-2020a)^3=(-2a)^3=-8a^3(1)$

Mặt khác:

$8(x-y)^2(y-z)=8(2018a-2019a)^2(2019a-2020a)=8a^2.(-a)=-8a^3(2)$

Từ $(1); (2)$ ta có đpcm.

Anh ơi em nghĩ phải lả \(+\frac{1}{x+y+z}\)thì mới đúng ạ

sửa đề \(M=\frac{x^2+1}{x}+\frac{y^2+1}{y}+\frac{z^2+1}{z}+\frac{1}{x+y+z}\)

                                giải

Áp dụng bđt cô si cho 3 số dương \(x,y,z\)ta có:

\(\hept{\begin{cases}x^2+1\ge2\sqrt{x^2}=2x\\y^2+1\ge2\sqrt{y^2}=2y\\z^2+1\ge2\sqrt{z^2}=2z\end{cases}}\)

\(\Rightarrow\frac{x^2+1}{x}\ge2;\frac{y^2+1}{y}\ge2;\frac{z^2+1}{z}\ge2\)(1)

Áp dụng bđt bunhiacopxki ta có:

\(\left(x+y+z\right)^2\le\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow\left(x+y+z\right)^2\le3^2\)

Mà \(x,y,z\)nguyên dương

\(\Rightarrow x+y+z\le3\)

\(\Rightarrow\frac{1}{x+y+z}\ge\frac{1}{3}\left(2\right)\)

Lấy (1) + (2) ta được:

\(M\ge2+2+2+\frac{1}{3}\)

\(\Rightarrow M\ge\frac{19}{3}\)

Dấu"="xảy ra \(\Leftrightarrow x=y=z\)

Lê Tài Bảo Châu Đề bài ko sai.

\(M=\frac{x^2+1}{x}+\frac{y^2+1}{y}+\frac{z^2+1}{z}-\frac{1}{x+y+z}\)

Theo ĐL Cool Kid đz luôn có \(\frac{1}{a+b+c}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\Rightarrow M\ge x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(\Rightarrow M\ge x+y+z+\frac{8}{9x}+\frac{8}{9y}+\frac{8}{9z}\)

Có BĐT :\(x+\frac{8}{9x}\ge\frac{x^2+33}{18}\Leftrightarrow.......\Leftrightarrow\left(x-1\right)^2\left(16-x\right)\ge0\left(true\right)\)

Tương tự cộng vế theo vế thì \(M\ge\frac{x^2+y^2+z^2+99}{18}=\frac{17}{3}\)

Dấu "=" xảy ra tại \(x=y=z=1\)

/\(2020\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{y^2+z^2}+\dfrac{1}{x^2+y^2}\right)ápdụngBDT\)

\(\dfrac{1}{x^2+y^2}+\dfrac{1}{y^2+z^2}+\dfrac{1}{x^2+z^2}\ge\dfrac{9}{2\left(x^2+y^2+z^2\right)}=\dfrac{9}{2\cdot2020}\)

\(ápdụngBĐTcosi\)

\(x^3+y^3+z^3\ge3xyz\)

\(\)=> VP\(\ge\) 9/2

\(\dfrac{x}{2017}=\dfrac{y}{2018}=\dfrac{z}{2019}=k\\ \Rightarrow\left\{{}\begin{matrix}x=2017k\\y=2018k\\z=2019k\end{matrix}\right.\)

\(4\left(x-y\right)\left(y-z\right)=4\left(2017k-2018k\right)\left(2018k-2019k\right)=4\left(-k\right)\left(-k\right)=4k^2=\left(2k\right)^2=\left(2019k-2017k\right)^2=\left(z-x\right)^2\left(ĐPCM\right)\)

Lời giải:
\(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{x+y-z}\Leftrightarrow \frac{x+y}{xy}=\frac{1}{z}+\frac{1}{x+y-z}=\frac{x+y}{z(x+y-z)}\)

\(\Leftrightarrow (x+y)(\frac{1}{xy}-\frac{1}{z(x+y-z)})=0\)

\(\Leftrightarrow (x+y).\frac{z(x+y-z)-xy}{xyz(x+y-z)}=0\)

\(\Leftrightarrow (x+y).\frac{(z-x)(y-z)}{xyz(x+y-z)}=0\)

\(\Leftrightarrow (x+y)(z-x)(y-z)=0\)

Xét các TH sau:

TH1: $x+y=0$. TH này loại do ĐKXĐ $x,y>0$
TH2: $z-x=0\Leftrightarrow z=x$

$\Leftrightarrow \frac{1}{y}=\frac{2020}{2021}$

\(M=\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{y}}=\frac{2}{\sqrt{y}}=2\sqrt{\frac{2020}{2021}}\)

TH3: $y-z=0$ tương tự TH2, ta có \(M=2\sqrt{\frac{2020}{2021}}\)

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